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Homework Statement
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.
Homework Equations
W = μmg * d
W = .5mvf2-.5mvi2
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/s2 * 3.01 m = 6.869 J
6.869 J = .5(.0262 kg)(298 m/s)2-.5(.0262 kg)Vi2
= 298.87 m/s
However, my concept is likely wrong because that's not the right answer. I would appreciate any help.