A block of mass 5 kg moves along the x-direction subject to the force $$F = (-20x + 10)$$ N, with the value of $$x$$ in metre. At time $$t = 0$$ s, it is at rest at position $$x = 1$$ m. The position and momentum of the block at $$t = (\pi/4)$$ s are

The force on the block is position-dependent: $$F(x)= -20x+10$$.

Rewrite it in the standard spring form:
$$F(x)= -20x+10 = -20\bigl(x-0.5\bigr)$$

Hence it behaves like a spring of effective force constant $$k=20\text{ N/m}$$ whose natural (equilibrium) position is
$$x_0 = 0.5\text{ m}$$ because $$F(x_0)=0$$.

For a mass $$m=5\text{ kg}$$ attached to such a “spring”, the angular frequency of simple harmonic motion is
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{20}{5}} = 2\text{ rad s}^{-1}$$.

Define the displacement from equilibrium as $$y = x - x_0$$. Then the SHM equation is $$\frac{d^2y}{dt^2}+ \omega^{2}y = 0$$, whose general solution is
$$y(t)=A\cos(\omega t + \phi)$$.

Initial conditions (given at $$t=0$$):
Position $$x(0)=1\text{ m}\;\Rightarrow\;y(0)=1-0.5 = 0.5\text{ m}$$
Velocity $$v(0)=0\;\Rightarrow\;\frac{dy}{dt}\Bigl|_{t=0}=0$$.

Using $$y(0)=A\cos\phi$$ and $$v(0)=-A\omega\sin\phi$$:

• $$-A\omega\sin\phi = 0 \;\Rightarrow\; \sin\phi = 0 \;\Rightarrow\; \phi = 0 \text{ or } \pi$$.
• With $$\phi = 0$$, $$A = y(0)/\cos 0 = 0.5\text{ m}$$ (if $$\phi=\pi$$ the cosine is $$-1$$ giving a negative displacement, which contradicts $$y(0)=+0.5\text{ m}$$).

Thus
$$y(t)=0.5\cos(2t)$$ and hence
$$x(t)=y(t)+x_0 = 0.5\cos(2t)+0.5$$.

Evaluate at $$t=\frac{\pi}{4}\text{ s}$$:

• Angle $$2t = 2\left(\frac{\pi}{4}\right)=\frac{\pi}{2}$$.
• Position $$x\!\left(\frac{\pi}{4}\right)=0.5\cos\!\frac{\pi}{2}+0.5=0+0.5=0.5\text{ m}$$.

Velocity is the time derivative of position:
$$v(t) = \frac{dx}{dt}= -0.5\cdot 2 \sin(2t)= -\sin(2t)$$.

At $$t=\frac{\pi}{4}\text{ s}$$, $$\sin\!\frac{\pi}{2}=1$$, so
$$v\!\left(\frac{\pi}{4}\right)= -1\text{ m/s}$$.

Momentum $$p = m v = 5\text{ kg}\times(-1\text{ m/s}) = -5\text{ kg m/s}$$.

Therefore, at $$t = \frac{\pi}{4}\text{ s}$$ the block is at $$x = 0.5\text{ m}$$ with momentum $$p = -5\text{ kg m/s}$$.

Option C which is: $$0.5\text{ m},\;-5\text{ kg m/s}$$.