First slide

Second law

Question

A bullet of mass 0.05 kg moving with a speed of 80 m/s enters a wooden block and is stopped after a distance of 0.40 m. The average resistive force exerted by the block on the bullet is:

Easy

A

300 N
B

20 N
C

400 N
D

40 N

Solution

$F_{av} = {ma}_{av} = m (\frac{v^{2}}{2 s}) = (0 .05) (\frac{80 \times 80}{2 \times 0 .40}) = 400 N$

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