A 5.00 g bullet is fired horizontally into a 2.00 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 0.40 m along the surface before stopping.
What is the initial speed of the bullet?
I don't know how the coefficient of kinetic energy comes to play here... can anyone help please? Thanks in advance.
2007-01-08 11:18:25 · 2 answers · asked by abc123 1 in Science & Mathematics ➔ Physics
You use the coefficient of knietic energy to help determan how much force was applied to the block via the bullet to cause the block to move 0.40 meters. I can't remeber the quations right now, but you work backwards, I belive. Takeing the distance traveled and the coefficent of friction to find the force.
2007-01-08 11:36:12 · answer #1 · answered by Weston 3 · 0⤊ 1⤋
You don't mean coefficient of kinetic energy, it's coefficient of kinetic friction. The amount of work friction did in bringing the block and bullet to a stop is
W=Ff*0.4m. Where Ff=mu*m*g.
That amount of work equals the kinetic energy the block had immediately after impact.
W = KE = (total mass/2)*v^2 Solve for v where v is the velocity of the pair stuck together.
Now use conservation of momentum to determine the velocity of bullet.
2007-01-08 12:36:11 · answer #2 · answered by sojsail 7 · 1⤊ 0⤋