Laws of Motion

From Newton's second law of motion,

$m a = m \dfrac{d^2 x}{d t^2} = F_0 \cos(\dfrac{\pi t}{2})$

Integrating with respect to time, and keeping in mind that the initial velocity is zero,

$\dfrac{dx}{dt} = \dfrac{F_0}{m} . \dfrac{2}{\pi} \sin(\dfrac{\pi t}{2})$

Note that velocity will be zero for the first time at $t = 2$ .

Next, we integrate with respect to time again to obtain the position,

$x(t) - x_0 = -\dfrac{F_0}{m} (\dfrac{2}{\pi})^2 \cos(\dfrac{\pi t}{2}) \Big|_0^t = -\dfrac{F_0}{m} (\dfrac{2}{\pi})^2 ( \cos(\dfrac{\pi t}{2}) - 1)$

Therefore, the displacement of the particle, at $t = 2$ is

$x(2) - x_0 = -\dfrac{F_0}{m} \dfrac{4}{\pi^2} (-1 - 1) = \dfrac{ 8 F_0 }{m \pi^2 }$

Hence, $k = 8$ .

$\begin{aligned} F & = F_0 \cos \left(\frac {\pi t}2\right) & \small \color{#3D99F6} \text{By Newton's second law} \\ m \color{#3D99F6}\ddot s & = F_0 \cos \left(\frac {\pi t}2\right) & \small \color{#3D99F6} \text{where }\ddot s = \frac {d^2s}{dt^2} \text{ is the acceleration of the particle.} \\ \implies \ddot s(t) & = \frac {F_0}m \cos \left(\frac {\pi t}2\right) \\ \color{#3D99F6} \dot s(t) & = \int \frac {F_0}m \cos \left(\frac {\pi t}2\right) dt & \small \color{#3D99F6} \text{where }\dot s = \frac {ds}{dt} \text{ is the velocity of the particle.} \\ & = \frac {2F_0}{m\pi} \sin \left(\frac {\pi t}2\right) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \dot s (0) & = C = \color{#3D99F6} 0 & \small \color{#3D99F6} \text{Initial velocity is 0.} \\ \implies \dot s(t) & = \frac {2F_0}{m\pi} \sin \left(\frac {\pi t}2\right) \\ \color{#3D99F6} s(t) & = \int \frac {2F_0}{m\pi} \sin \left(\frac {\pi t}2\right) dt & \small \color{#3D99F6} \text{where }s(t) \text{ is the displacement of the particle.} \\ & = - \frac {4F_0}{m\pi^2} \cos \left(\frac {\pi t}2\right) + C \\ s(0) & = - \frac {4F_0}{m\pi^2} + C = 0 \\ C & = \frac {4F_0}{m\pi^2} \\ \implies s(t) & = \frac {4F_0}{m\pi^2} \left(1-\cos \left(\frac {\pi t}2\right)\right) \end{aligned}$

The particle first stops, when $\dot s(t) = \dfrac {2F_0}{m\pi} \sin \left(\dfrac {\pi t}2\right) = 0$ . That is $t=2$ .

The distance traveled by the particle then $s(2) = \dfrac {4F_0}{m\pi^2} \left(1-\cos \pi \right) = \dfrac {8F_0}{m\pi^2}$

$\implies k = \boxed{8}$