Answer
(a) The work done on the particle from x = 0 to x = L is $\frac{1}{2}mv_0^2$
(b) The work done on the particle from x = 0 to x = 2L is zero.
Work Step by Step
(a) The work done on the particle is equal to the change in kinetic energy.
We can find the speed at x = 0
$v_x = v_0~sin(\frac{\pi~x}{2L})$
$v_x = v_0~sin(\frac{\pi~(0)}{2L})$
$v_x = 0$
We can find the speed at x = L
$v_x = v_0~sin(\frac{\pi~x}{2L})$
$v_x = v_0~sin(\frac{\pi~(L)}{2L})$
$v_x = v_0~sin(\frac{\pi~}{2})$
$v_x = v_0$
We can find the change in kinetic energy.
$\Delta KE = KE_2-KE_1$
$\Delta KE = \frac{1}{2}mv_0^2-0$
$\Delta KE = \frac{1}{2}mv_0^2$
The work done on the particle from x = 0 to x = L is $\frac{1}{2}mv_0^2$
(b) The work done on the particle is equal to the change in kinetic energy.
We can find the speed at x = 0
$v_x = v_0~sin(\frac{\pi~x}{2L})$
$v_x = v_0~sin(\frac{\pi~(0)}{2L})$
$v_x = 0$
We can find the speed at x = 2L
$v_x = v_0~sin(\frac{\pi~x}{2L})$
$v_x = v_0~sin(\frac{\pi~(2L)}{2L})$
$v_x = v_0~sin(\pi)$
$v_x = 0$
We can find the change in kinetic energy.
$\Delta KE = KE_2-KE_1$
$\Delta KE = 0-0$
$\Delta KE = 0$
The work done on the particle from x = 0 to x = 2L is zero.
