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 = .5mvf2-.5mvi2 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/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.