Problem 42E
A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping. What was the initial speed of the bullet?
Step-by-Step Solution:
Solution 42E Step 1: Mass of the bullet is, m1= 5 gram = 0.005 kg . Mass of the wooden box is, m = 1.20 kg. 2 Coefficient of kinetic friction, = 0.20. k Distance covered, s = 0.310 m. Step 2: Work done by the frictional force is, W fric F fric s = k s = 0.20 × Mg × 0.310 = 0.20 × (m + m ) ×19.8 2 0.310 = 0.20 × (0.005 + 1.20) × 9.8 × 0.310 = 0.732 J .
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Chapter 8, Problem 42E is Solved
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