# Physics GRE GR 1777, Problem 019 Solution

## 19. Classical Mechanics (Work-Energy Principle)

##### Solution
From the work-energy theorem,

$W = \Delta K = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2$$W = \Delta K = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2$

Since the work done by the box is

$W = F \cdot d = F \cdot (5 m)$$W = F \cdot d = F \cdot (5 m)$

and the change of the kinetic energy is

$\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 = \frac{1}{2} (10 kg) \cdot [ (2 m/s)^2 - (1 m/s)^2] = 15 kg \cdot m^2/s^2$$\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 = \frac{1}{2} (10 kg) \cdot [ (2 m/s)^2 - (1 m/s)^2] = 15 kg \cdot m^2/s^2$

Therefore, The force is

$F = 3 kg \cdot m/s^2 = 3N$$F = 3 kg \cdot m/s^2 = 3N$

(C) 3N
https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle