A body of mass 2 kg slides down with an acceleration of 3 m/s$$^2$$ on a rough inclined plane having a slope of 30$$^\circ$$. The external force required to take the same body up the plane with the same acceleration will be: (g = 10 m/s$$^2$$)

We begin with the motion of the block when it is allowed to slide down the rough incline without any external pull. Let the mass be $$m = 2\ \text{kg}$$, the angle of the incline be $$\theta = 30^{\circ}$$ and the downward acceleration be $$a = 3\ \text{m s}^{-2}$$. The acceleration due to gravity is given to be $$g = 10\ \text{m s}^{-2}$$.

The component of the weight along the plane is obtained from the trigonometric resolution

$$W_{\parallel}= mg\sin\theta.$$

Substituting the numerical values,

$$W_{\parallel}= 2 \times 10 \times \sin 30^{\circ}= 20 \times \dfrac12 = 10\ \text{N}.$$

The normal reaction is

$$N = mg\cos\theta = 2 \times 10 \times \cos 30^{\circ} = 20 \times \dfrac{\sqrt3}{2}=10\sqrt3\ \text{N}\approx 17.32\ \text{N}.$$

Let the coefficient of kinetic friction be $$\mu_k$$. The frictional force (opposing the motion) is therefore

$$f = \mu_k N = \mu_k \; mg \cos\theta.$$

Along the line of the slope, the downward component of weight drives the motion while friction opposes it. Newton’s second law in the downward direction gives

$$mg\sin\theta - f = ma.$$

Substituting $$f = \mu_k mg\cos\theta$$ we obtain

$$mg\sin\theta - \mu_k mg\cos\theta = ma.$$

Dividing every term by $$m$$ to simplify,

$$g\sin\theta - \mu_k g\cos\theta = a,$$

and therefore

$$\mu_k g\cos\theta = g\sin\theta - a.$$

Now substitute the numbers:

$$\mu_k \;(10)\cos 30^{\circ}= 10\sin 30^{\circ}-3.$$

$$\mu_k \;(10)\times\dfrac{\sqrt3}{2}= 10\times\dfrac12 -3.$$

$$5\sqrt3\,\mu_k = 5-3 = 2.$$

$$\mu_k = \dfrac{2}{5\sqrt3}=\dfrac{2\sqrt3}{15}\approx 0.231.$$

Knowing $$\mu_k$$ makes it possible to find the actual frictional force:

$$f = \mu_k N = \mu_k mg\cos\theta = \left(\dfrac{2}{5\sqrt3}\right)\bigl(2\times10\cos30^{\circ}\bigr) = \dfrac{2}{5\sqrt3}\bigl(20\times\dfrac{\sqrt3}{2}\bigr)=4\ \text{N}.$$

Next we tackle the second part of the problem. Now the block has to move up the same incline with an upward acceleration of the same magnitude, viz. $$a = 3\ \text{m s}^{-2}$$. We apply an external force $$F$$ parallel to the plane and directed upward. Choosing the upward direction as positive, we write Newton’s second law once again.

The forces acting along the plane while the block is accelerating upward are

• External pull $$F$$ upward (positive).
• Component of weight $$mg\sin\theta = 10\ \text{N}$$ downward (negative).
• Friction, which always opposes the motion, is now downward. Its magnitude is already known to be $$4\ \text{N}$$, so it is taken as $$-4\ \text{N}$$ in the upward sign convention.

Applying $$\sum F = ma$$ with the upward direction taken as positive, we therefore have

$$F - mg\sin\theta - f = m a.$$

Substituting the numerical values,

$$F - 10 - 4 = 2 \times 3.$$

$$F - 14 = 6.$$

$$F = 6 + 14 = 20\ \text{N}.$$

Thus an external force of $$20\ \text{N}$$ parallel to the incline and directed upward is necessary to make the block climb the plane with an acceleration of $$3\ \text{m s}^{-2}$$.

Hence, the correct answer is Option D.