A 20.0 gram bullet is shot horizontally into a 2.00 kg block and quickly becomes imbedded in the block. After the impact, the block slides along a rough horizontal surface until it comes to a stop after moving 85 cm. The coefficient of kinetic friction between the block and the surface is μk = 0.45. How much energy is dissipated in the block during the impact?

Question

A 20.0 gram bullet is shot horizontally into a 2.00 kg block and quickly becomes imbedded in the block. After the impact, the block slides along a rough horizontal surface until it comes to a stop after moving 85 cm. The coefficient of kinetic friction between the block and the surface is μk = 0.45.                                                                   How much energy is dissipated in the block during the impact?

anonymous · Answer

Stopping distance = 
d = V^2 / 2 * u * g

d = distance = 0.85m
V = velocity of object before friction applied
u = kinetic friction coeff = 0.45
g = gravity = 9.81 m/s^2

d = V^2 / (2 * u * g)
0.85m = V^2 / 2 * 0.45 * 9.81 m/s^2
V^2 = (2 * 0.45 * 9.81 m/s^2) * 0.85m
V = sq rt [(2 * 0.45 * 9.81 m/s^2) * 0.85m]
Velocity of block prior to deceleration by friction:
V = 2.73 m/s

Now take kinetic energy knowing the block had to be going 2.73 m/s before it slid to stop:
KE = 1/2 mv^2
KE = 1/2 (0.02kg + 2kg) (2.73 m/s)^2
KE = 7.5 Joules of energy lost in impact

anonymous · Answer

Thank you so much!  I totally did not understand what they were asking.