(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 26.2 g bullet is fired horizontally into a

1.32 kg wooden block resting on a horizon-

tal surface (μ = 0.173). The bullet goes

through the block and comes out with a speed

of 298 m/s.

If the block travels 3.01 m before coming to

rest, what was the initial speed of the bullet?

The acceleration of gravity is 9.8 m/s2 .

Answer in units of m/s.

2. Relevant equations

W = μmg * d

W = .5mv_{f}^{2}-.5mv_{i}^{2}

3. The attempt at a solution

So I thought that the kinetic energy transferred by the bullet to the block would be the force of friction of the block.

W = .173 * (1.32 kg + .0262 kg) * 9.8 m/s^{2}* 3.01 m = 6.869 J

6.869 J = .5(.0262 kg)(298 m/s)^{2}-.5(.0262 kg)V_{i}^{2}

= 298.87 m/s

However, my concept is likely wrong because that's not the right answer. I would appreciate any help.

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# Bullet strikes block on friction surface

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