In a ballistics test, a 28-g bullet pierces a sand bag that is 30${~cm}$ thick. If the initial bullet velocity was 55${~m} / \mathrm { s }$ and it emerged from the sandbag moving at 18${~m} / \mathrm { s } \text {, }$ what was the magnitude of the friction force (assuming it to be constant) that the bullet experienced while it traveled through the bag?
A) 130 N
B) 38 N
C) 13 N
D) 1.3 N

Question

Quizplus · Accepted Answer

The correct answer is A.The change in kinetic energy of the bullet is:$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$$$\Delta K = \frac{1}{2}(0.028 	ext{ kg})(18 	ext{ m/s})^2 - \frac{1}{2}(0.028 	ext{ kg})(55 	ext{ m/s})^2$$$$\Delta K = -154 	ext{ J}$$The work done by the friction force is:$$W_f = F_fd$$$$F_fd = \Delta K$$$$F_d = \frac{\Delta K}{d} = \frac{-154 	ext{ J}}{0.3 	ext{ m}}$$$$F_d = 513 	ext{ N}$$

In a Ballistics Test, a 28-G Bullet Pierces a Sand cm{~cm} cm

In a Ballistics Test, a 28-G Bullet Pierces a Sand ${~cm}$