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

Question

Quizplus · Accepted Answer

Explanation:The change in kinetic energy of the bullet is:$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$$$\Delta K = \frac{1}{2}(0.028 \mathrm{~kg})(18 \mathrm{~m/s})^2 - \frac{1}{2}(0.028 \mathrm{~kg})(55 \mathrm{~m/s})^2$$$$\Delta K = -154 \mathrm{~J}$$The work done by the friction force is:$$W_f = F_fd$$$$F_fd = \Delta K$$$$F_d = \frac{\Delta K}{d} = \frac{-154 \mathrm{~J}}{0.30 \mathrm{~m}}$$$$F_d = \boxed{513 \mathrm{~N}}$$

In a Ballistics Test, a 28-G Bullet Pierces a Sand 30 cm30 \mathrm {~cm}30 cm

In a Ballistics Test, a 28-G Bullet Pierces a Sand $30 \mathrm {~cm}$