Chapter 7: Work, Energy, and Simple Machines

Concept involved : When the child starts from rest at the top of the slide, gravity converts the whole loss of gravitational potential energy into kinetic energy (we neglect friction and air-resistance, as is usually assumed in Class 9 problems).

Symbols used

• $$m$$ – mass of the child
• $$h$$ – vertical height of the slide (top to bottom)
• $$g$$ – acceleration due to gravity ( $$9.8\,\mathrm{m\,s^{-2}}$$ )
• $$u$$ – speed of the child at the top (starts from rest, so $$u = 0$$)
• $$v$$ – speed of the child at the bottom (what we have to find)

Step 1 : Write the two forms of mechanical energy

Potential energy at the top  $$U_{\text{top}} = mgh$$
Kinetic energy at the bottom  $$K_{\text{bottom}} = \dfrac12 m v^2$$

Step 2 : Apply conservation of mechanical energy

Because no external work is done on the child–slide system,

$$mgh = \dfrac12 m v^2$$

Step 3 : Solve algebraically for $$v$$

First cancel the common factor $$m$$ on both sides :

$$gh = \dfrac12 v^2$$

Multiply both sides by $$2$$

$$2gh = v^2$$

Take the square root

\[v = \sqrt{2gh}\]

Step 4 : Substitute the numerical values

For the blue slide the textbook gives a height of $$h = 3.0\,\mathrm{m}$$ (all slides in the figure have the same height).

$$v = \sqrt{2 \times 9.8\,\mathrm{m\,s^{-2}} \times 3.0\,\mathrm{m}}$$

$$v = \sqrt{58.8}\,\mathrm{m\,s^{-1}}$$

\[v \approx 7.7\,\mathrm{m\,s^{-1}}\]

Final result : The magnitude of the child’s velocity at the bottom of the blue slide is about $$7.7\,\mathrm{m\,s^{-1}}$$ (≈ 28 km h^-1). Notice that the shape of the slide has no effect on this value; only the vertical height matters.

Given : Two children, A and B, of different masses m₁ and m₂, start from rest at the same position on the top of a smooth slide whose bottom is at a vertical depth h below the starting point.

The question is whether their speeds at the bottom will be the same.

1. Write the forms of mechanical energy.

• Gravitational potential energy (P.E.) of a child of mass m at height h: $$U = mgh$$.
• Kinetic energy (K.E.) when the child reaches the bottom with speed v: $$K = \tfrac12 m v^2$$.

2. Apply conservation of mechanical energy.

The slide is taken as smooth (no friction), and air resistance is neglected. Therefore, mechanical energy is conserved:

Potential energy at the top = Kinetic energy at the bottom.

For any child on the slide,

$$mgh = \tfrac12 m v^2$$.

3. Solve for the speed v.

Cancel the common factor m from both sides:

$$gh = \tfrac12 v^2 \;\;\;\Longrightarrow\;\;\; v^2 = 2gh$$.

Taking the positive square-root (speed is positive), we get the key result

4. Interpret the result.

• The expression (1) contains only g (acceleration due to gravity) and h (vertical height).
• It is independent of mass; the symbols m₁ and m₂ never appear in the final formula.

5. Conclusion.

Since both children start from the same height and the mass does not affect the final speed given by equation (1), both children will reach the bottom of the slide with the same velocity, provided friction and air resistance are negligible.

Extra note (beyond ideal conditions) : In the real world, a heavier child may experience slightly less fractional loss of energy due to friction, so the speeds could differ a little. However, according to the ideal laws of mechanics taught at this level, the two velocities are identical.

Given: Every slide begins at the same vertical height h. The child starts from rest and, as stated in the chapter discussion, we neglect friction and air resistance, so only gravity does work.

Step 1 – Apply conservation of mechanical energy
The total mechanical energy at the top (T) must equal that at the bottom (B):

• At T: potential energy $$U_T = m g h$$, kinetic energy $$K_T = 0$$ (child is at rest).
• At B: potential energy $$U_B = 0$$ (ground is the reference level), kinetic energy $$K_B = \tfrac12 m v^2$$.

Hence

gives

Step 2 – Solve for the speed

Step 3 – Compare the slides
The final speed $$v$$ depends only on the vertical drop $$h$$ and the acceleration due to gravity $$g$$ — not on the slide’s shape or length. Because every slide starts from the same height, each gives exactly the same magnitude of velocity at the bottom.

Conclusion: None of the slides produces a larger speed; the child reaches the bottom of every slide with the same velocity.

(If substantial friction were present, the shorter, steeper slide would give a slightly higher speed, but the textbook question assumes the ideal friction-free case.)

Concept recalled

For a constant force the mechanical work is defined as

$$W = \vec F \cdot \vec s = F s \cos \theta,$$

where $$\theta$$ is the angle between the (average) force $$\vec F$$ applied by the agent and the displacement $$\vec s$$ of the body on which the force acts.

• If $$0^\circ \le \theta < 90^\circ$$, $$\cos \theta > 0$$ → $$W > 0$$ (positive work).
• If $$\theta = 90^\circ$$, $$\cos 90^\circ = 0$$ → $$W = 0$$ (zero work).
• If $$90^\circ < \theta \le 180^\circ$$, $$\cos \theta < 0$$ → $$W < 0$$ (negative work).

Situation 1: Lifting the dumb-bell upward

• The girl exerts an upward muscular force $$\vec F_{\text{girl}}$$ to overcome the weight $$\vec W = m\vec g$$.
• The displacement of the dumb-bell during lifting is also upward. Hence $$\theta = 0^\circ$$ between $$\vec F_{\text{girl}}$$ and $$\vec s$$.

Therefore

$$W_{\text{girl, up}} = F_{\text{girl}}\, s \cos 0^\circ = F_{\text{girl}}\, s ( +1 ) > 0.$$

So the girl does positive work on the dumb-bell while lifting it.

Situation 2: Lowering the dumb-bell slowly (i.e. almost at constant speed)

• Once again she exerts an upward force $$\vec F_{\text{girl}}$$, but of the same magnitude as the weight so that the net force is practically zero and the dumb-bell comes down slowly without acceleration.
• Now the displacement $$\vec s$$ of the dumb-bell is downward.
• The angle between her force (upward) and the displacement (downward) is $$\theta = 180^\circ$$.

Hence

$$W_{\text{girl, down}} = F_{\text{girl}}\, s \cos 180^\circ = F_{\text{girl}}\, s ( -1 ) < 0.$$

Thus the girl does negative work on the dumb-bell while lowering it. (At the same time gravity does an equal amount of positive work.)

Result summarised

• Positive work: during the upward lift.
• Negative work: during the controlled downward motion.

Given data

• Magnitude of force applied by the goalkeeper:
  $$F = 200 \text{ N}$$
• Backward displacement of the hand (same as the displacement of the ball while being stopped):
  $$s = 15 \text{ cm} = 0.15 \text{ m}$$

Step 1 — Identify the angle between force and displacement

The goalkeeper's force is exerted opposite to the direction in which the ball is moving. Hence the angle between the applied force (towards the player) and the ball’s displacement (still forward, though slowing down) is $$\theta = 180^{\circ}$$.

Step 2 — Write the work-done formula

For a constant force at an angle $$\theta$$ to the displacement,

$$W = F\,s\,\cos\theta$$

Step 3 — Substitute the known values

$$W = (200\,\text{N})(0.15\,\text{m})\cos 180^{\circ}$$

Since $$\cos 180^{\circ} = -1$$,

$$W = (200)(0.15)(-1)\,\text{J}$$

Step 4 — Evaluate

$$W = -30\,\text{J}$$

The negative sign shows that the goalkeeper’s hand removed energy from the ball (work done against its motion), thereby stopping it.

Step 1 — Recall the definition of mechanical work
When a constant force $$\vec F$$ acts on a body and the centre of mass of the body undergoes a displacement $$\vec s$$, the work done by the force is defined as
\[ W = \vec F \cdot \vec s = F s \cos \theta \]
where $$\theta$$ is the angle between $$\vec F$$ and $$\vec s$$.

Step 2 — Identify the force and displacement for the barbell
The weightlifter exerts an upward force $$F = mg$$ on the barbell to balance its weight. While she is holding it steady, the barbell does not move upward or downward. Hence its displacement is
$$ s = 0 $$.

Step 3 — Compute the work done
Using the work formula with $$s = 0$$:
$$ W = F s \cos \theta = F (0) \cos \theta = 0 \text{ J}. $$

Step 4 — Interpretation
Because the barbell’s displacement is zero, the mechanical work done on it is zero, even though the weightlifter’s muscles consume energy internally to keep the force applied.

Therefore, while holding the barbell steady, the weightlifter does no mechanical work on the barbell.

Problem restated
The stack of coins slides forward on a horizontal rough surface (Fig. 6.13c). We have to decide whether the work done by the frictional force exerted by the surface on the coins is positive, negative or zero.

1. Identify the directions
   • Displacement of the coins, $$\vec{s}$$: forward.
   • Kinetic friction on the coins, $$\vec{F}_f$$: exactly opposite to the motion (backward), because friction always opposes relative motion.

2. Use the definition of work
   By definition, the work done by a constant force is
   $$W = \vec{F}_f \cdot \vec{s} = F_f\,s\cos\theta,$$ where $$\theta$$ is the angle between $$\vec{F}_f$$ and $$\vec{s}$$.

3. Substitute the angle
   Since the two vectors point in opposite directions, $$\theta = 180^{\circ}$$. For this angle $$\cos 180^{\circ} = -1$$.

4. Sign of the work
   \[ W = F_f\,s\cos 180^{\circ} = -F_f\,s < 0 \]
   Hence the numerical value of the work is negative.

5. Conclusion
   The work done by friction on the moving stack of coins is negative. (This negative work removes kinetic energy and eventually brings the coins to rest.)

Given information
The striker is flicked by the player, hits an intermediate (white) coin, which in turn strikes the required black coin and pockets it. We neglect air resistance and sliding friction on the board for the very short time‐intervals of interest, but we do keep track of the small losses to sound and heat that inevitably accompany every collision.

Step 1 Who actually does external work?

• The only agent that supplies energy from outside the system of coins is the player’s finger/hand. While flicking, the muscles exert a force $$\vec F$$ on the striker through a displacement $$\vec s$$, so the work done is

\[W = \vec F\, \cdot \vec s > 0\]

• This positive work converts the player’s chemical energy of muscles into the kinetic energy of the striker. No one else (the intermediate coin, the black coin, or the pockets) performs external work; afterwards the coins merely exchange the energy already present in the system.

Step 2 Energy changes at every collision

1. Finger → Striker
   Chemical energy (muscles) $$\longrightarrow$$ Kinetic energy of striker $$K_s = \tfrac12 m_s v_s^2$$.
2. Striker → Intermediate (white) coin
   During the (slightly inelastic) impact the striker exerts an impulsive force on the coin.
   • Work done by striker on the coin is the loss of its own kinetic energy:
   $$\Delta K_s = -W_{s\,\text{on}\,w}$$,
   $$\Delta K_w = +W_{s\,\text{on}\,w}$$.
   • Result: a large part of $$K_s$$ becomes kinetic energy of the white coin; a small part is dissipated as sound + heat.
3. White coin → Black coin
   In the next collision the now-moving white coin does work on the stationary black coin.
   $$K_w \longrightarrow K_b$$ (kinetic of black coin) + sound + heat.
   The white coin slows down almost to rest; the black coin shoots towards the pocket.
4. Black coin → Pocket / board felt
   As the black coin slides and finally drops into the pocket its kinetic energy is completely converted into:
   • Heat produced by sliding friction with the board felt.
   • Sound made when it touches the base of the pocket.
   There is no useful mechanical energy left once the coin comes to rest.

Summary of energy flow

Conclusion
The player is the only one who does external mechanical work. Every subsequent contact merely redistributes the kinetic energy already present, each time bleeding off a small fraction as sound and heat until the black coin finally stops in the pocket.

Given concept

Muscular energy originates from the chemical energy of the food stored in our body. Whenever we do mechanical work (for example, pedalling a bicycle), this muscular energy is transformed into other forms of energy.

Application to the present situation

1. To set the bicycle and rider in motion
      • The chief useful output is the kinetic energy of translation of the bicycle–rider system.
      • The rotating parts (wheels, crank, chain sprockets) also gain rotational kinetic energy.
      Therefore a major part of the muscular energy becomes mechanical (kinetic) energy: $$E_{\text{muscle}} \longrightarrow E_{\text{kinetic}}\;.$$
2. To overcome resistive forces
      Even on a level road the rider must work continuously because of
   • rolling friction between tyre and road,
   • internal friction in the chain, bearings, gears, and
   • air resistance (drag).
      The work done against these forces finally appears as heat in the tyres, road surface, bicycle parts and the surrounding air: $$E_{\text{muscle}} \longrightarrow E_{\text{heat}}\;.$$
3. Minor energy channels
      A very small fraction is converted into sound produced by the chain, gears or the tyres rubbing the road: $$E_{\text{muscle}} \longrightarrow E_{\text{sound}}.$$

Summary of energy transformations

Thus, while pedalling on a flat road, the chemical energy stored in our muscles is mainly converted into:

• mechanical (kinetic) energy of the bicycle and its moving parts,
• heat energy (due to friction and air resistance), and
• sound energy (a negligible share).

Kinetic energy of a body is given by the formula $$K = \tfrac{1}{2} m v^{2}$$, where $$m$$ is the mass of the body and $$v$$ is its speed.

Step 1 – Original kinetic energy
If the initial speed of the vehicle is $$v$$, its initial kinetic energy is

Step 2 – Kinetic energy after doubling the speed
When the speed is doubled, the new speed becomes $$2v$$. Substituting $$2v$$ for $$v$$ in the formula,

$$K_2 = \tfrac{1}{2} m (2v)^{2}$$

Simplify the square:

$$K_2 = \tfrac{1}{2} m (4v^{2})$$

Factor the numerical coefficient:

$$K_2 = 4 \left( \tfrac{1}{2} m v^{2} \right)$$

But $$\tfrac{1}{2} m v^{2} = K_1$$, therefore

Conclusion
Doubling the speed makes the kinetic energy four times its original value.

Given data

• Mass of the cricket ball: $$m = 0.2\,\text{kg}$$
• Speed of the delivery: $$v = 154.8\,\text{km\,h}^{-1}$$

Step 1: Convert the speed to metres per second (SI unit)

We know $$1\,\text{km\,h}^{-1} = \tfrac{1000\,\text{m}}{3600\,\text{s}} = \tfrac{5}{18}\,\text{m\,s}^{-1}$$.

So

$$v = 154.8 \times \frac{5}{18}\,\text{m\,s}^{-1}$$

First divide by 18:

$$\frac{154.8}{18} = 8.6$$

Then multiply by 5:

$$v = 8.6 \times 5 = 43.0\,\text{m\,s}^{-1}$$

Step 2: Write the formula for kinetic energy

For an object of mass $$m$$ moving with speed $$v$$, the kinetic energy $$E_{\text{k}}$$ is

$$E_{\text{k}} = \tfrac{1}{2} m v^{2}$$

Step 3: Substitute the numerical values

$$E_{\text{k}} = \tfrac{1}{2} \times 0.2\,\text{kg} \times (43.0\,\text{m\,s}^{-1})^{2}$$

First square the velocity:

$$v^{2} = 43.0^{2} = 1849$$

Multiply by the mass:

$$m v^{2} = 0.2 \times 1849 = 369.8$$

Now take half of this product:

$$E_{\text{k}} = \tfrac{1}{2} \times 369.8 = 184.9$$

Step 4: State the final answer with appropriate significant figures

\[E_{\text{k}} \;\approx\; 1.85 \times 10^{2}\, \text{J}\]

Thus, the kinetic energy of the ball at the moment of delivery is approximately $$185\,\text{joules}$$.

Given data

• Mass of the jet, $$m = 15000 \; \mathrm{kg}$$
• Retarding (backward) force, $$F = 367500 \; \mathrm{N}$$
• Stopping distance, $$s = 100 \; \mathrm{m}$$
• Final velocity after being stopped, $$v_f = 0 \; \mathrm{m\,s^{-1}}$$

We have to find the initial velocity $$v_i$$ of the aircraft just before the hook engaged the wire.

Method : Work–energy theorem

The work done by the constant backward (retarding) force equals the change in kinetic energy of the aircraft.

Work done by the force:

Because the force is opposite to the direction of motion, the angle between the force and the displacement is $$180^{\circ}$$, so $$\cos 180^{\circ} = -1$$.

$$W = F \; s \; \cos 180^{\circ} = -F s$$

Change in kinetic energy:

$$\Delta K = K_f - K_i = \frac12 m v_f^{2} - \frac12 m v_i^{2} = 0 - \frac12 m v_i^{2} = -\frac12 m v_i^{2}$$

According to the work–energy theorem,

$$W = \Delta K$$

Substituting the expressions for $$W$$ and $$\Delta K$$:

$$-F s = -\frac12 m v_i^{2}$$

Cancel the negative signs on both sides:

$$F s = \frac12 m v_i^{2}$$

Solve for $$v_i$$:

$$v_i^{2} = \frac{2 F s}{m}$$

Now plug in the given values:

$$v_i^{2} = \frac{2 \times 367500 \; \mathrm{N} \times 100 \; \mathrm{m}}{15000 \; \mathrm{kg}}$$

First, multiply the numerator:

$$2 \times 367500 = 735000$$

$$735000 \times 100 = 73500000$$

So,

$$v_i^{2} = \frac{73500000}{15000}$$

Divide:

$$v_i^{2} = 4900$$

Take square root:

$$v_i = \sqrt{4900} = 70 \; \mathrm{m\,s^{-1}}$$

Therefore, the aircraft was moving at $$70 \; \mathrm{m\,s^{-1}}$$ just before the wire caught the hook.

Step 1 &vert; Write the expression for kinetic energy

For any object of mass $$m$$ moving with speed $$v$$, the kinetic energy is

Step 2 &vert; Write kinetic energy for each object

• Object A: mass $$m$$, speed $$v_A$$
      $$K_A = \tfrac12 m v_A^2$$
• Object B: mass $$4m$$, speed $$v_B$$
      $$K_B = \tfrac12 (4m) v_B^2$$

Step 3 &vert; Set the kinetic energies equal

It is given that the two kinetic energies are the same:

$$K_A = K_B$$

Substitute the expressions just written:

$$\tfrac12 m v_A^2 = \tfrac12 (4m) v_B^2$$

Step 4 &vert; Simplify the equality

Cancel the common factor $$\tfrac12 m$$ from both sides:

$$v_A^2 = 4 v_B^2$$

Step 5 &vert; Take the square root

$$v_A = 2 v_B$$ (Only the positive root is taken because speed is a magnitude.)

Step 6 &vert; Write the required ratio

Thus, the magnitude of the velocity of object A is twice that of object B.

Given : An object of mass m is moving with constant (uniform) velocity $$\vec v$$.

Concept used : The kinetic energy (K.E.) of a body is defined as

\[K = \frac{1}{2} m v^{2}\]

where v is the magnitude of the velocity.

Explanation :

• When the motion is with constant velocity, its magnitude $$v$$ remains the same at every instant of time.
• The mass m of the object is, of course, fixed.
• Since both factors that appear in the formula for kinetic energy – mass m and speed v – stay unchanged, the numerical value of $$K$$ does not vary as the object changes its position along the path.

Hence, the kinetic energy is constant; it does not depend on the position of the object so long as the velocity remains constant.

Given data

• Mass of the cricket ball: $$m = 200 \text{ g}$$
• Maximum height above the ground: $$h = 10 \text{ m}$$
• Acceleration due to gravity: $$g = 10 \mathrm{ m\,s^{-2}}$$

Step 1: Convert mass to SI units

$$200 \text{ g} = \frac{200}{1000} \text{ kg} = 0.2 \text{ kg}$$

Step 2: Recall the formula for gravitational potential energy

The gravitational potential energy $$U$$ at height $$h$$ is

$$U = mgh.$$

Step 3: Substitute the known values

$$U = (0.2 \text{ kg})(10 \mathrm{ m\,s^{-2}})(10 \text{ m}).$$

Step 4: Calculate

$$U = 0.2 \times 10 \times 10 = 20 \text{ J}.$$

Result

The ball’s gravitational potential energy at the highest point is

\[U = 20\;\text{J}\]

Known relation for gravitational potential energy near Earth’s surface

For an object of mass $$m$$ kept at height $$h$$ (measured from an arbitrary reference level), the gravitational potential energy is

\[U = m g h\] where $$g$$ is the acceleration due to gravity (≈ 9.8 m s⁻²).

(i) Motion with constant velocity in the horizontal direction

1. The object moves parallel to the Earth’s surface, so its vertical coordinate (its height $$h$$ above the chosen reference level) remains the same:
       $$h_2 = h_1$$.
2. Initial potential energy: $$U_1 = m g h_1$$.
3. Final potential energy: $$U_2 = m g h_2 = m g h_1$$.
4. Change in potential energy:
       $$\Delta U = U_2 - U_1 = m g h_1 - m g h_1 = 0$$.

Therefore, while the object moves horizontally (with any speed), its gravitational potential energy does not change.

(ii) Motion in the vertical direction (object gradually raised)

1. Let the object be lifted from height $$h_1$$ to a greater height $$h_2$$, so $$h_2 > h_1$$.
2. Initial potential energy: $$U_1 = m g h_1$$.
3. Final potential energy: $$U_2 = m g h_2$$.
4. Change in potential energy:
       $$\Delta U = U_2 - U_1 = m g (h_2 - h_1)$$.
5. Since $$h_2 - h_1 > 0$$, we get $$\Delta U > 0$$; i.e. the potential energy increases in direct proportion to the rise in height.

Conclusion

• Horizontal motion at constant level ⟹ no change in gravitational potential energy.
• Vertical upward motion ⟹ gravitational potential energy increases by $$m g \Delta h$$.

Given data

• Height of the top of the slide above the bottom: $$h$$
• Mass of the child: $$m$$ (it will cancel out, but we keep it during the derivation)
• Initial speed at the top: the child starts from rest, so $$u = 0$$
• The slide is considered smooth (negligible friction and air resistance), therefore mechanical energy is conserved.

Step 1 – Write the mechanical energy at the top

At the top the child is at height $$h$$. Hence

• Potential energy (PE)
$$\text{PE}_{\text{top}} = m g h$$

• Kinetic energy (KE)
Because $$u = 0$$,

$$\text{KE}_{\text{top}} = \tfrac12 m u^2 = \tfrac12 m (0)^2 = 0$$

• Total mechanical energy at the top

$$E_{\text{top}} = \text{PE}_{\text{top}} + \text{KE}_{\text{top}} = m g h + 0 = m g h$$

Step 2 – Write the mechanical energy at the bottom

At the bottom the child’s height is zero, so

• Potential energy
$$\text{PE}_{\text{bottom}} = m g (0) = 0$$

• Let the unknown speed at the bottom be $$v$$. Then the kinetic energy is

$$\text{KE}_{\text{bottom}} = \tfrac12 m v^2$$

• Total mechanical energy at the bottom

$$E_{\text{bottom}} = \text{PE}_{\text{bottom}} + \text{KE}_{\text{bottom}} = 0 + \tfrac12 m v^2 = \tfrac12 m v^2$$

Step 3 – Apply conservation of mechanical energy

Because the slide is smooth, no energy is lost to friction, so

$$E_{\text{top}} = E_{\text{bottom}}$$

Substitute the two energies:

$$m g h = \tfrac12 m v^2$$

Step 4 – Solve for $$v$$

First cancel the common factor $$m$$ (mass does not affect the result):

$$g h = \tfrac12 v^2$$

Multiply both sides by $$2$$:

$$2 g h = v^2$$

Now take the square root of both sides (speed is positive):

\[v = \sqrt{2 g h}\]

The magnitude of the child’s velocity on reaching the bottom depends only on $$g$$ and the vertical height $$h$$ of the slide.

1. Convert the truck’s speed to SI units

$$v = 72\;\mathrm{km\,h^{-1}} = 72 \times \frac{1000\,\mathrm{m}}{3600\,\mathrm{s}} = 20\;\mathrm{m\,s^{-1}}$$

2. Initial kinetic energy of the truck

$$K_i = \tfrac12 m v^2 = \tfrac12 (10000\;\mathrm{kg})(20\;\mathrm{m\,s^{-1}})^{2} = 2.0\times10^{6}\;\mathrm{J}$$

3. Forces opposing the motion on the slope

• Component of truck’s weight along the incline:
  $$W_{\parallel} = mg \sin\theta = (10000)(10)\sin30^{\circ} = 50000\;\mathrm{N}$$
• Retarding force exerted by sand:
  $$F_s = 50000\;\mathrm{N}$$

Thus the total retarding force is

$$F_{\text{total}} = W_{\parallel} + F_s = 50000\;\mathrm{N} + 50000\;\mathrm{N} = 100000\;\mathrm{N}$$

4. Energy required to bring the truck to rest

The work done by the total retarding force over a distance $$s$$ along the ramp equals the initial kinetic energy:

$$F_{\text{total}} \; s = K_i$$

Substituting the numbers,

$$100000\;\mathrm{N}\;\times s = 2.0\times10^{6}\;\mathrm{J}$$

Hence

5. Check: vertical rise

With $$\sin30^{\circ}=\tfrac12$$ the truck climbs

$$h = s\sin\theta = 20\times\tfrac12 = 10\;\mathrm{m}$$

which is consistent with the energy used against gravity plus sand.

Minimum length of the escape ramp

Given: A ball of mass $$m$$ is allowed to fall freely from a height $$h$$ above the ground (Fig. 7.19). The ground level is chosen as the zero level for gravitational potential energy.

Step 1 – Initial mechanical energy (at height h)

• Initial kinetic energy: the ball is released from rest, so $$K_i = \tfrac12 m u^2 = 0$$ because $$u = 0$$.
• Initial potential energy: $$U_i = m g h$$ (distance of centre of the ball above the ground is $$h$$).

Hence the total (mechanical) energy when the ball is released is

$$E_i = K_i + U_i = 0 + m g h = m g h.$$

Step 2 – Speed just before hitting the ground

The ball falls freely under gravity, so we use the kinematic relation

$$v^2 - u^2 = 2 a s,$$

where $$u = 0$$, $$a = g$$ (downwards) and $$s = h$$ (distance fallen). Therefore

$$v^2 = 2 g h \;\;\;\Rightarrow\;\; v = \sqrt{2 g h}.$$

Step 3 – Mechanical energy just before impact

• Kinetic energy just before impact:

  $$K_f = \tfrac12 m v^2 = \tfrac12 m (2 g h) = m g h.$$

• Potential energy at the ground (zero level): $$U_f = 0.$$

The total mechanical energy at this instant is

$$E_f = K_f + U_f = m g h + 0.$$

Step 4 – Comparison with the initial energy

The initial and final totals are identical:

\[E_f = m g h = E_i.\]

Thus, even just before the ball strikes the ground, its mechanical energy is still $$m g h$$, confirming the conservation of mechanical energy for free fall.

Given situation – A smooth-looking but real track in a science park has the shape shown in Fig. 7.22 (a succession of hills and valleys). A steel ball is released from the highest point A. The marked positions are

• A – the starting (highest) point
• B – the first lowest point, just after A
• C – the next crest after B
• D, E – still later crests

We have to discuss how gravitational potential energy (PE) and kinetic energy (KE) vary at A, B, C and why the later crests (C, D, E…) are lower than the previous ones.

Choose the horizontal floor of the park as the zero-level of potential energy. Let

• mass of the ball  = $$m$$
• $$g$$ = acceleration due to gravity
• heights of the points above the reference level = $$h_A,\,h_B,\,h_C\,(\dots)$$

1. Energy at the top A

The ball is simply released, so initial speed $$u=0$$.

Potential energy at A: $$PE_A = m g h_A$$
Kinetic energy at A: $$KE_A = \dfrac12 m u^2 = 0$$

Total mechanical energy at A is therefore

\[E_A = PE_A + KE_A = m g h_A\]

2. Energy at the bottom B

While rolling down from A to B the ball loses height, so some PE converts into KE.

Height at B (above the chosen reference) is $$h_B$$, so

Potential energy at B: $$PE_B = m g h_B$$

If friction and air resistance are ignored for the moment, the loss in PE equals the gain in KE:

$$KE_B = m g (h_A - h_B)$$

Hence at B, KE is maximum because the height is minimum, whereas PE is minimum.

3. Energy at the next crest C

The ball now climbs up from B to C. Height increases from $$h_B$$ to $$h_C$$, so some of the kinetic energy is reconverted into potential energy.

Potential energy at C: $$PE_C = m g h_C$$

Available mechanical energy just before reaching C (again neglecting friction) would still be

$$E = m g h_A$$ (the same as at A).

Therefore, ideal kinetic energy at C would be

$$KE_C = m g (h_A - h_C)$$

This tells us:

• If there were no dissipative forces, the ball would rise to exactly the original height, i.e. $$h_C = h_A$$ and $$KE_C = 0$$. It would momentarily come to rest before rolling down again, just like an ideal pendulum.
• In the real track, however, some energy is lost as heat and sound because of rolling friction at the axle, deformation of the rails, air resistance, etc. Let the energy lost in going from A to C be $$E_{\text{loss}}$$.

Then the actual mechanical energy available at C is

$$E_C = m g h_A - E_{\text{loss}}$$

Equating this to $$PE_C + KE_C$$ we get

$$m g h_C + KE_C = m g h_A - E_{\text{loss}}$$

Since $$E_{\text{loss}} > 0$$, the same total energy cannot be maintained, which implies $$h_C < h_A$$. In words, the ball cannot climb back to its original height.

4. Why the successive crests keep getting lower (C, D, E…)

Each time the ball rolls along the track it experiences:

• Rolling friction between the ball and the rails
• Internal friction within the ball and the track due to small deformations
• Air resistance, especially when the speed is high near the bottom

All these forces are non-conservative. They convert a part of the mechanical energy into heat and sound, which the ball-track system cannot recover. Consequently the mechanical energy keeps decreasing:

\[E_A > E_C > E_D > E_E > \dots\]

Because potential energy at any crest is $$m g h$$, a fall in total mechanical energy inevitably shows up as a lower attainable height. Therefore every new high point (C, D, E, …) is lower than the immediately preceding one, until finally the ball stops somewhere in the first valley when all its initial mechanical energy has been dissipated.

Summary of energy changes

The reduction in the height of successive crests is indeed due to the energy irreversibly lost to friction and air resistance.

Given data

• Mass lifted: $$m = 75\,\text{kg}$$
• Height raised: $$h = 2\,\text{m}$$
• Time taken: $$t = 5\,\text{s}$$
• Acceleration due to gravity: $$g = 9.8\,\mathrm{m\,s^{-2}}$$

Step 1: Work done against gravity

$$W = mgh$$

$$W = 75\,\text{kg} \times 9.8\,\mathrm{m\,s^{-2}} \times 2\,\text{m}$$

$$W = 75 \times 19.6\,\text{J}$$

$$W = 1470\,\text{J}$$

Step 2: Power required

Power is the rate of doing work:

$$P = \dfrac{W}{t}$$

$$P = \dfrac{1470\,\text{J}}{5\,\text{s}}$$

$$P = 294\,\text{W}$$

Conclusion

The weightlifter needs a power of about $$3.0 \times 10^{2}\,\text{W}$$ (approximately 300 W) to lift the mass in the given time.

Given data

• Mass of the car, $$m = 1000 \;\mathrm{kg}$$
• Initial speed, $$u = 0$$ (starts from rest)
• Final speed, $$v = 72\;\mathrm{km\,h^{-1}}$$
• Time taken, $$t = 10\;\mathrm{s}$$

Step 1 – Convert the final speed to SI units

$$v = 72\;\mathrm{km\,h^{-1}} = 72 \times \frac{1000}{3600}\;\mathrm{m\,s^{-1}} = 20\;\mathrm{m\,s^{-1}}$$

Step 2 – Find the change in kinetic energy

The car starts from rest, so its initial kinetic energy is zero. The final kinetic energy is

$$K = \tfrac12 m v^{2}$$

Substituting the values:

$$K = \tfrac12 \times 1000\;\mathrm{kg} \times (20\;\mathrm{m\,s^{-1}})^{2}$$

$$K = \tfrac12 \times 1000 \times 400$$

$$K = 200\,000\;\mathrm{J}$$

This is the work $$W$$ done by the engine during the 10 s start, because on a level road work done $$=\,$$ gain in kinetic energy.

Step 3 – Calculate the power of the engine

Power is the rate of doing work:

$$P = \frac{W}{t}$$

$$P = \frac{200\,000\;\mathrm{J}}{10\;\mathrm{s}}$$

$$P = 20\,000\;\mathrm{W}$$

$$P = 20\;\mathrm{kW}$$

Result

The engine must supply a power of

\[P = 20\;\text{kW}\]

Given data

• Vertical height to be climbed (step height), $$h = 30\;\text{cm}$$.
• Horizontal reach (width of the ramp on the floor), $$b = 40\;\text{cm}$$.

For an inclined plane we need its actual length (the sloping surface) to calculate the ideal mechanical advantage (IMA).

1. Find the length of the ramp

The ramp, the height and the horizontal reach form a right‑angled triangle, so by Pythagoras’ theorem:

$$l = \sqrt{h^{2}+b^{2}}$$

$$l = \sqrt{(30\;\text{cm})^{2} + (40\;\text{cm})^{2}}$$

$$l = \sqrt{900 + 1600}\;\text{cm}$$

$$l = \sqrt{2500}\;\text{cm}$$

$$l = 50\;\text{cm}$$

2. Calculate the mechanical advantage

For an ideal (frictionless) inclined plane:

$$\text{Mechanical Advantage (IMA)} = \dfrac{\text{Length of slope}}{\text{Vertical height}}$$

$$\text{IMA} = \dfrac{l}{h} = \dfrac{50\;\text{cm}}{30\;\text{cm}}$$

$$\text{IMA} = \dfrac{5}{3} \approx 1.67$$

Therefore, the ramp multiplies the applied effort by a factor of about 1.67.

The hillside road is nothing but a big inclined plane. To see why a long, gently-sloping incline is preferred to a short, steep one, compare the force a vehicle must exert in the two cases.

Let the vehicle of mass $$m$$ finally reach a height $$h$$ above the starting point.

1. Work needed is fixed by the height, not by the path

The gravitational potential-energy gained is

Equation (1) shows that the total work that has to be done against gravity is the same whether we go straight up or wind around.

2. Force needed on an inclined plane

Suppose the road rises the same height $$h$$ but is made as an incline of length $$l$$, making an angle $$\theta$$ with the horizontal (Fig. 4.26). The component of the vehicle’s weight parallel to the road is

$$F = m g \sin\theta.$$

Because $$\sin\theta = h / l$$, the required uphill pulling force becomes

$$F = m g \dfrac{h}{l}. $$

Longer road ⇒ larger $$l$$ ⇒ smaller $$F$$.

3. Mechanical advantage

For an inclined plane the mechanical advantage is

$$\text{M.A.} = \dfrac{\text{load}}{\text{effort}} = \dfrac{m g}{F} = \dfrac{l}{h}. $$

The larger the ratio $$l/h$$ (that is, the gentler and longer the slope), the less effort the engine has to supply at each instant. While the engine works for a longer distance, it never has to exceed its comfortable force (or power) limit, preventing stalling, overheating, and skidding; passengers also enjoy a safer, smoother ride.

Therefore, roads on hills are made to wind gently instead of rising abruptly, so that the force (and power) required from vehicles at any moment becomes small enough for ordinary engines to manage, even though the total work $$mgh$$ remains unchanged.

Given situation

• Height to be reached  =  $$h$$.
• Weight of the climber  =  $$mg$$.
• Two possible paths:

1. Vertical ladder  →  distance climbed $$=h$$.
2. Inclined ladder: length of ladder $$=L$$, makes angle $$\theta$$ with the horizontal; the same vertical height $$h$$ is reached.

1. Forces acting while climbing very slowly (no acceleration)

The only external force the climber has to balance is the component of the weight acting along the ladder.

Resolve the weight $$mg$$ along (∥) and perpendicular (⟂) to the ladder:

Along the ladder: $$mg\sin\theta$$   [because the angle between $$mg$$ and the normal to the plane is $$\theta$$].

2. Force a climber must apply

To move up very slowly (quasi-static) the net force along the ladder is zero, so the climber’s upward pull must be

For a vertical ladder $$\theta = 90^{\circ}$$, therefore $$F_{\text{vertical}} = mg\sin 90^{\circ} = mg$$.

For any inclined ladder $$0^{\circ} < \theta < 90^{\circ}$$, so $$\sin\theta < 1$$ and

$$F_{\text{inclined}} = mg\sin\theta < mg = F_{\text{vertical}}.$$

3. Work done in each case

Work is the product of the force actually applied and the distance through which that force acts.

(i) Vertical climb:

$$W_v = mg \times h = mgh$$.

(ii) Inclined climb:

\[ W_i = F \times L = \bigl(mg\sin\theta\bigr) L. \]

But for the right-triangle formed by the ladder, $$\sin\theta = \dfrac{h}{L}$$ ⇒ $$L = \dfrac{h}{\sin\theta}$$.

Substituting,

The work needed is the same in both cases; energy is conserved.

4. Why the inclined ladder feels easier

• The required force is smaller, eqn. (1).
• Human effort is usually limited by the maximum force our muscles can exert rather than by the total energy we can expend over a few seconds. A smaller force therefore feels “easier”, even though it has to be exerted over a longer distance.
• This is exactly how an inclined plane (a simple machine) provides mechanical advantage $$\text{M.A.}=\dfrac{\text{load}}{\text{effort}} = \dfrac{mg}{mg\sin\theta} = \dfrac{1}{\sin\theta}=\dfrac{L}{h}>1.$$(Bigger M.A. ⇒ smaller effort).

Conclusion

Climbing an inclined ladder requires the same total work but a smaller force; hence it is felt to be easier than climbing a vertical ladder.

Given data

Distances of the four seats from the fulcrum C:

• Left-hand side: seat A at $$AC = 2\,\text{m}$$, seat B at $$BC = 1\,\text{m}$$
• Right-hand side: seat D at $$DC = 1\,\text{m}$$, seat E at $$EC = 2\,\text{m}$$

Masses of the two children:

$$m_1 = 15\,\text{kg}, \qquad m_2 = 30\,\text{kg}$$

Principle of moments (law of the lever)

The seesaw is in equilibrium when the algebraic sum of the turning moments about the fulcrum is zero. Taking clockwise moments as positive and anticlockwise moments as negative, we need

$$m_1 g \times r_1 = m_2 g \times r_2$$

where $$r_1$$ and $$r_2$$ are the respective perpendicular distances of the two children from C, on opposite sides.

Choosing suitable seats

Because $$m_2 = 2 m_1$$ (the second child is twice as heavy), the heavier child should sit at half the distance from C compared with the lighter child:

$$r_2 = \dfrac{m_1}{m_2}\,r_1 = \dfrac{1}{2}\,r_1$$

Checking the available seat positions:

We therefore place

• the lighter child (15 kg) on a seat 2 m from C (either A or E);
• the heavier child (30 kg) on the diametrically opposite seat 1 m from C (either D or B).

Verification

Example choice: 15 kg at A, 30 kg at D.

Anticlockwise moment: $$m_1 g r_1 = 15g \times 2 = 30g$$
Clockwise moment: $$m_2 g r_2 = 30g \times 1 = 30g$$

Since the two moments are equal and opposite, the seesaw is balanced.

Hence, any of the following symmetric arrangements gives equilibrium:

• 15 kg on A and 30 kg on D, or
• 15 kg on E and 30 kg on B.

Concept involved : A spoon used in the manner shown in Fig. 7.35 behaves like a first-class lever.

The rim of the can works as the fulcrum (O). The blade of the spoon just under the lid exerts an upward force on the lid (this is the load, L). The force you apply at the handle is the effort, E.

Moments (turning effects) about the fulcrum

For rotational equilibrium of the lever, the clockwise moment of the load must equal the anticlockwise moment of the effort:

$$ F_L\,l_L = F_E\,l_E $$

where

• $$F_L$$ = force needed to lift the lid (load)
• $$F_E$$ = force applied by your hand (effort)
• $$l_L$$ = load arm = perpendicular distance of the load from fulcrum (very small, only the thickness of the spoon blade)
• $$l_E$$ = effort arm = distance of your hand from fulcrum (length of the handle, several centimetres)

Re-arranging,

$$ F_E = \dfrac{l_L}{l_E}\,F_L $$

Because $$l_E \gg l_L$$, the ratio $$\dfrac{l_L}{l_E}$$ is a small fraction, so the required effort $$F_E$$ is much smaller than the actual force needed to raise the lid directly.

Mechanical advantage

The mechanical advantage (M.A.) of a lever is

$$ \text{M.A.}=\dfrac{F_L}{F_E}=\dfrac{l_E}{l_L}\;\;$$

Since $$l_E \gg l_L$$, we get a large M.A. Hence the spoon multiplies the force of your hand, making it easier to open the can.

Energy viewpoint

The work you do is $$W=F_E\,l_E$$ and the work done on the lid is $$F_L\,l_L$$. From the equality of moments, these two works are equal, so energy is conserved. The spoon does not reduce the work; it only allows the same work to be done with a smaller force acting through a larger distance.

Conclusion : A spoon provides a long effort arm and a short load arm, giving a large mechanical advantage. Therefore a small force exerted at the handle produces a much larger force at the lid, making it easier to open the can.

Scissors are examples of class-I levers (fulcrum between effort and load). To understand why we cut hard objects close to the fulcrum, analyse the moment (torque) condition for a lever.

1. Label the three key points.
   
   • Fulcrum F: the rivet/pin about which the blades turn.
   • Effort E: the force your fingers apply on the handles.
   • Load L: the resistance offered by the material being cut, acting at the point on the blades where the material is placed.
2. Torque (moment) balance.
   

   For rotational equilibrium, clockwise moment = anticlockwise moment:

   $$E \times L_e = L \times L_l$$
   where
   $$L_e$$ = effort arm (distance F‒E) and $$L_l$$ = load arm (distance F‒L).

3. Mechanical advantage.
   

   Dividing by $$E$$ gives

   $$\text{M.A.}=\dfrac{L}{E}=\dfrac{L_e}{L_l}$$

   The smaller the load arm $$L_l$$, the larger the mechanical advantage.

4. Shifting the load closer to the fulcrum.
   

   When you slide the material toward F, $$L_l$$ decreases while $$L_e$$ (length of the handle) stays the same. Hence

   $$\text{M.A.}\;{\Large\uparrow}\qquad(\because L_l \downarrow)$$

   For the same muscular effort $$E$$ the load that can now be overcome is

   $$L = \dfrac{E\,L_e}{L_l}$$

   Because $$L_l$$ is smaller, $$L$$ becomes much larger; the blades therefore press on the material with a greater force.

5. Result.
   

   To cut a hard object you push it right next to the rivet so that the shortened load arm gives the scissors a higher mechanical advantage and a larger cutting force, allowing the hard object to be sheared with the same hand effort.

Step 1 — Condition for endless motion

For a wheel or any machine to keep rotating without ever slowing down, its total mechanical energy (kinetic + potential) must remain constant:

This requires the net work done by all non-conservative (resistive) forces to be zero.

Step 2 — Resistive forces in every real machine

• Friction at the axle, joints and bearings
• Air resistance on the moving surfaces
• Internal deformation of the parts, producing heat and sound

All these forces oppose the motion at every point, so the angle between the resistive force $$\vec F_{\text{res}}$$ and the displacement $$\vec s$$ is $$180^{\circ}$$. Their work is therefore negative:

Step 3 — Apply the work–energy theorem to one revolution

Let the wheel of mass $$m$$ begin a revolution with speed $$v_0$$. Its initial kinetic energy is

During that revolution the resistive forces perform the negative work $$W_{\text{res}}$$ given by Eq. (2). The work–energy theorem states

If no external agent supplies positive work, then $$W_{\text{net}} = W_{\text{res}}$$, and substituting from Eq. (2) gives

Since $$F_{\text{res}}\, s$$ is positive, Eq. (5) shows that $$K_1$$ is smaller than $$K_0$$. The kinetic energy after the revolution is less than it was before, so the speed drops.

Step 4 — Repeat over many revolutions

The same loss occurs in every subsequent revolution. After $$n$$ revolutions,

Every $$W_i$$ is negative, so the kinetic energy steadily decreases. Eventually $$K_n \to 0$$, the speed reaches zero, and the machine comes to rest.

Step 5 — Energy is conserved overall

The mechanical energy that the machine appears to lose is not destroyed; it is converted into internal energy (heat) of the bearings, the air and the machine's parts, with a small fraction radiated as sound. The law of conservation of energy is therefore obeyed — only the mechanical share of the energy decreases.

Conclusion

No real machine can be free of friction and air resistance. These resistive forces always do negative work on the moving parts, continuously draining their mechanical energy into heat and sound. The kinetic energy keeps falling until the motion stops. Hence a perpetual-motion machine is impossible.

Given: A ball is thrown vertically upward. We examine its physical quantities at the highest point of its flight (air resistance neglected).

1. Force on the ball.
   Throughout the motion, the only significant force on the ball is its weight $$W = mg$$, directed vertically downward. Even at the highest point — where the ball is momentarily at rest — gravity continues to act, so the force does not become zero.
   Conclusion: Statement (i) is false.
2. Acceleration of the ball.
   By Newton's second law, \[ a = \dfrac{F_{\text{net}}}{m} = \dfrac{mg}{m} = g \] directed downward. The acceleration due to gravity $$g \approx 9.8\,\mathrm{m\,s^{-2}}$$ is the same at every point of the trajectory, including the highest one.
   Conclusion: Statement (ii) is false.
3. Kinetic energy at the highest point.
   The velocity becomes zero at the top: $$v = 0$$. Therefore \[ K = \tfrac{1}{2} m v^{2} = 0 \] Conclusion: Statement (iii) is true.
4. Potential energy at the highest point.
   Gravitational potential energy is $$U = mgh$$, where $$h$$ is the height measured from the point of projection. The ball is at its greatest height, so $$U$$ is maximum.
   Conclusion: Statement (iv) is true.

Final result: The correct statements are (iii) and (iv).

Given data and symbols

• Mass lifted: $$m$$
• Height of one floor: $$h$$
• First lift: to the 10^th floor, so vertical rise $$10h$$ in time $$t$$ (the statement says “a certain time”; call it $$t$$).
• Second lift: to the 20^th floor, so vertical rise $$20h$$ in time $$2t$$ (“double the time”).
• Acceleration due to gravity: $$g$$.

1. Work / energy for each lift

Change in gravitational potential energy = work done by the crane.

First lift:

$$W_1 = m g (10h)$$

Second lift:

$$W_2 = m g (20h)$$

$$\Rightarrow \; W_2 = 2\,m g (10h) = 2W_1$$

Extra energy required

$$W_2 - W_1 = 2W_1 - W_1 = W_1 = m g (10h)$$

Thus the second operation needs twice as much energy as the first; the additional energy equals the whole of the first lift’s energy (100 % more).

2. Power for each lift

Average power = work ÷ time.

First lift:

$$P_1 = \dfrac{W_1}{t} = \dfrac{m g (10h)}{t}$$

Second lift:

$$P_2 = \dfrac{W_2}{2t} = \dfrac{m g (20h)}{2t}$$

Simplify:

$$P_2 = \dfrac{20}{2}\,\dfrac{m g h}{t} = 10\,\dfrac{m g h}{t}= \dfrac{m g (10h)}{t}=P_1$$

Extra power required

$$P_2 - P_1 = 0$$

The crane must supply the same power in both cases; there is no increase in power requirement.

Result

• Energy for the 20^th-floor lift is double that for the 10^th-floor lift; the “extra” energy equals $$m g (10h)$$ (a 100 % increase).
• Power remains unchanged; no additional power is needed.

Step 1 – Identify what is being lifted and through what height
Let the mass of the flag (together with the short length of rope that actually rises) be $$m$$. The flag is pulled from the ground to the top of a pole of height $$h$$. The vertical distance through which the centre of gravity of the flag is raised is therefore $$h$$.

Step 2 – Write the expression for the gravitational potential energy gained
When an object of mass $$m$$ is lifted vertically through a height $$h$$ against gravity, the increase in its gravitational potential energy is

where $$g$$ is the acceleration due to gravity (approximately $$9.8\,\text{m\,s}^{-2}$$). This increase in potential energy is exactly the mechanical work that must be done by the person (through the pulley) on the flag, provided we neglect friction in the pulley and the weight of the rope.

Therefore, the energy required depends only on

• the mass $$m$$ of the flag (and any part of the rope that is lifted),
• the vertical height $$h$$ through which it is raised, and
• the constant $$g$$ (same at a given place on the Earth).

Whether the flag is raised slowly or quickly does not change $$m$$, $$g$$ or $$h$$, so the work done (and hence the energy required) remains exactly the same.

Step 3 – Effect of speed on power
Suppose the flag is raised at a uniform speed $$v$$. The time taken is

The rate at which work is done, i.e. the power, is

Thus power is directly proportional to the lifting speed $$v$$.

If the speed is doubled (from $$v$$ to $$2v$$), the time taken halves (from $$t$$ to $$t/2$$) and

So the power required becomes twice the original value, even though the total work done is unchanged.

Summary

• Energy (work) needed to hoist the flag: $$W = mgh$$  → depends only on $$m$$, $$g$$ and $$h$$.
• Raising it slowly or quickly: same work, because $$W$$ does not involve time or speed.
• Doubling the speed halves the time and therefore doubles the power requirement: $$P \propto v$$.

Step 1 – List the masses on each day

• Mass of the man: $$60 \text{ kg}$$
• Mass of the scooter: $$100 \text{ kg}$$
• Mass of the son: $$40 \text{ kg}$$ (only on the second day)

First day (man + scooter):
$$m_1 = 60 + 100 = 160 \;\text{kg}$$

Second day (man + son + scooter):
$$m_2 = 60 + 40 + 100 = 200 \;\text{kg}$$

Step 2 – Write the work (energy) required each day

The scooter starts from rest and reaches the same speed $$v$$ in both cases. The gain in kinetic energy on each day equals the work done by the engine, and this work comes entirely from burning fuel (no other losses are assumed).

First day:
$$W_1 = \tfrac12 m_1 v^2$$

Second day:
$$W_2 = \tfrac12 m_2 v^2$$

Step 3 – Relate fuel consumption to the work done

The chemical energy released from the fuel is directly proportional to the work done, so

$$F_1 : F_2 = W_1 : W_2$$

Substituting $$W_1$$ and $$W_2$$ gives

$$\displaystyle F_1 : F_2 = \tfrac12 m_1 v^2 : \tfrac12 m_2 v^2 = m_1 : m_2$$

Step 4 – Insert the numerical masses

$$F_1 : F_2 = 160 : 200 = 4 : 5$$

Conclusion

The scooter burns fuel in the ratio 4 : 5; that is, if it uses 4 units of fuel on the first day, it will use 5 units on the second day.

Principle involved — Law of the Lever

For a seesaw (a type I lever) to be in equilibrium, the algebraic sum of the turning moments (torques) about the fulcrum must vanish:

$$\sum \tau = 0 \implies W_C\,x_C = W_A\,x_A$$

where

• $$W_C$$ = weight of the child,
• $$W_A$$ = weight of the adult,
• $$x_C$$ = perpendicular distance of the child from the fulcrum,
• $$x_A$$ = perpendicular distance of the adult from the fulcrum.

Given data

The adult is twice as heavy as the child:

$$W_A = 2W_C$$

Substitute into the moment balance

$$W_C\,x_C = (2W_C)\,x_A$$

Cancel the common factor $$W_C$$ (it is non-zero):

$$x_C = 2x_A$$

Result

The child must sit at a distance twice that of the adult from the fulcrum.

For example, if the adult adjusts his sliding seat so that he is $$x=1\,\text{m}$$ from the fulcrum, the child must sit $$x_C = 2\,\text{m}$$ away on the opposite side to balance the seesaw.

What to draw

• Draw a horizontal plank pivoted at its centre on a triangular support (the fulcrum).
• Mark the left-hand seat (child) at a distance labelled $$x_C = 2x$$ from the fulcrum and draw a downward arrow labelled $$W_C$$.
• Mark the right-hand seat (adult) at a distance labelled $$x_A = x$$ from the fulcrum and draw a double-length downward arrow labelled $$W_A = 2W_C$$.
• Show the fulcrum exactly midway between the two riders.

This sketch makes it clear that the larger weight sits closer and the smaller weight sits farther so that the clockwise and anticlockwise moments are equal.

Given data
Mass of block $$m = 10.0\;\text{kg}$$
Initial kinetic energy at the origin $$K_{0}=180\;\text{J}$$

The force–displacement graph shown in Fig. 7.37 is a triangle that rises uniformly from 0 N at $$s = 0\;\text{m}$$ to 160 N at $$s = 2\;\text{m}$$ and then falls uniformly back to 0 N at $$s = 4\;\text{m}$$. (For the answer it is the area under the graph that matters, not its exact shape.)

1. Initial speed at 0 m

The kinetic-energy formula gives

$$K_{0}=\tfrac12 m v_{0}^{2}\;\;\Longrightarrow\;\;v_{0}=\sqrt{\dfrac{2K_{0}}{m}}$$

$$v_{0}=\sqrt{\dfrac{2\times 180\;\text{J}}{10\;\text{kg}}}=\sqrt{36}=6\;\text{m s}^{-1}$$

2. Work done by the variable force between 0 m and 4 m

The work is the shaded area under the graph.

• First triangle (0 m → 2 m): $$W_{1}=\tfrac12\times 2\;\text{m}\times 160\;\text{N}=160\;\text{J}$$
• Second triangle (2 m → 4 m): $$W_{2}=\tfrac12\times 2\;\text{m}\times 160\;\text{N}=160\;\text{J}$$

Total work

$$W=W_{1}+W_{2}=160\;\text{J}+160\;\text{J}=320\;\text{J}$$

3. Final speed at 4 m

Work–energy theorem: $$K_{4}=K_{0}+W$$

$$K_{4}=180\;\text{J}+320\;\text{J}=500\;\text{J}$$

The speed $$v_{4}$$ corresponding to this kinetic energy is

$$K_{4}=\tfrac12 m v_{4}^{2}\;\;\Longrightarrow\;\;v_{4}=\sqrt{\dfrac{2K_{4}}{m}}$$

$$v_{4}=\sqrt{\dfrac{2\times 500}{10}}=\sqrt{100}=10\;\text{m s}^{-1}$$

4. Does the block ever have negative acceleration?

Throughout the 4 m span the applied force is in the same direction as motion, so $$F\gt 0\;\Rightarrow\;a=F/m\gt0$$ at every point. Although the magnitude of the force (hence the acceleration) decreases after 2 m, it never becomes negative. Therefore, the block never experiences negative acceleration (retardation) during the given interval.

Given data

• Maximum height reached on Earth: $$h_E = 8 \;\text{m}$$
• Acceleration due to gravity on Earth: $$g_E$$ (symbolic)
• Acceleration due to gravity on Moon: $$g_M = \dfrac{g_E}{6}$$

Step 1: Initial speed on Earth

The kinematic relation for vertical motion is

$$v^2 = u^2 + 2 a s$$

At the highest point $$v = 0$$ and $$a = -g_E$$, so

$$0 = u^2 - 2 g_E h_E \;\Rightarrow\; u^2 = 2 g_E h_E \quad(1)$$

Step 2: Height on the Moon with the same speed

On the Moon the upward motion obeys

$$0 = u^2 - 2 g_M h_M \;\Rightarrow\; h_M = \dfrac{u^2}{2 g_M}$$

Because $$g_M = \dfrac{g_E}{6}$$,

$$h_M = \dfrac{u^2}{2 \bigl(\dfrac{g_E}{6}\bigr)} = \dfrac{6 u^2}{2 g_E} = 3\,\dfrac{u^2}{g_E} \quad(2)$$

Step 3: Substitute $$u^2$$ from (1)

Insert $$u^2 = 2 g_E h_E$$ into (2):

$$h_M = 3\,\dfrac{2 g_E h_E}{g_E} = 6 h_E$$

Step 4: Numerical value

$$h_M = 6 \times 8 \;\text{m} = 48 \;\text{m}$$

Result

The ball will rise to a height of about $$48\,\text{m}$$ from the surface of the Moon.

Step 1 – Read the potential–energy values from the graph

• At point O (starting point) the graph shows a potential energy of $$30\,\text{J}$$.​
• From the graph, the heights corresponding to the other three marked positions are:
    P : $$U_P = 20\,\text{J}$$,
    Q : $$U_Q = 0\,\text{J}$$,
    R : $$U_R = 30\,\text{J}$$.

Step 2 – Find the total mechanical energy of the ball

The ball is released from rest at O, so its kinetic energy there is zero.

$$E_{\text{total}} = U_O + K_O = 30\,\text{J} + 0 = 30\,\text{J}$$

Because the track is friction-less, this total mechanical energy remains the same everywhere on the track.

Step 3 – Relate kinetic energy and speed

At any point, $$K = \tfrac12\,m v^2$$, hence

$$v = \sqrt{\dfrac{2K}{m}}$$

The mass of the ball is $$m = 0.5\,\text{kg}$$.

Step 4 – Calculate speed at each point

Step 5 – State the results

The velocities of the 0.5 kg ball are therefore:

• at P  : $$v_P \approx 6.3\,\text{m s}^{-1}$$
• at Q  : $$v_Q \approx 11\,\text{m s}^{-1}$$
• at R  : $$v_R = 0\,\text{m s}^{-1}$$