A 2.0-gram bullet is shot into a tree stump. It enters at a speed of 3.00 x 104 cm/s and comes to rest after having penetrated 0.05 m in to the stump. What was the average force during the impact? Show all calculations leading to an answer.

Question

Physics

Miah

22 August, 09:41

A 2.0-gram bullet is shot into a tree stump. It enters at a speed of 3.00 x 104 cm/s and comes to rest after having penetrated 0.05 m in to the stump. What was the average force during the impact? Show all calculations leading to an answer.

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Answers (1)

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Iris Valenzuela · Answer 1

The stump does (negative) work on the bullet to slow it down:

Work = - force * distance

The work/energy theorem says that the change in the bullet's KE equals the total work done on the bullet.

change in KE = Work = - force * distance

The change in KE equals the final KE (zero) minus the initial KE (½m (v_initial) ²):

0 - ½m (v_initial) ² = - force * distance