A 3.0-g bullet traveling at 350 m/s hits a tree and slows down uniformly to a stop while penetrating a distance of 12 cm into the tree's trunk. What was the force exerted on the bullet in bringing it to rest?

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- Jul 12th 2006, 10:12 AM #1

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- Jul 12th 2006, 01:53 PM #2Originally Posted by
**Celia**

The work done on the bullet by the tree is going to be negative (consider the direction of the force and the displacement). I will assume an average force, so we may simply call the force constant. The final speed of the bullet is, of course, zero.

$\displaystyle W = - \bar F d = (1/2)mv^2 - (1/2)mv_0^2$

$\displaystyle \bar F = \frac{(1/2)mv^2}{d} = \frac{0.5 \cdot 0.003 \cdot (350)^2}{0.12} = 1531.25$

So the average force exerted by the tree over the 12 cm is about 1500 N.

-Dan