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A 9.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. the coefficient of kinetic friction between block and surface is 0.20. the bullet remains embedded in the block, which is observed to slide 0.340 m along the surface before stopping. part a what was the initial speed of the bullet? express your answer with the appropriate units.

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Demiurgos
We can start by observing what happens with energy in this situations. The bullet has some kinetic energy at the start and this kinetic energy is then used to do work against the friction acting on the wooden block.
[tex]\frac{m_bv_b^2}{2}=F_fL[/tex]
We need to figure out the force of friction to calculate the initial velocity of the bullet. 
We know that force of friction is equal to:
[tex]F_f=N\mu=Mg\mu[/tex]
The wooden block is resting on a horizontal surface, that means that normal force acting on it is the same as the gravitational force pulling it down. One more thing that we need to keep in mind is that bullet is embedded in the block. This means that bullet's weight also contributes to the total weight of the block.
[tex]F_f=(m_{wb}+m_b)g\mu[/tex]
[tex]\frac{m_bv_b^2}{2}=(m_{wb}+m_b)g\mu L\\ {m_bv_b^2}=2(m_{wb}+m_b)g\mu L\\ v_b=\sqrt{2(m_{wb}+m_b)g\mu L}/m_b\\[/tex]
When we plug in all the number we get:
[tex]v_b=141.12\frac{m}{s}[/tex]

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