work done on the block/bullet:
force*distance
(.046+1.27)*9.8*.28*16= 57.8Joules
KE of bullet=1/2 m v^2
57.8=1/2 * .046*v^2
solve for velocity v.
A 46 g bullet strikes and becomes embedded in a 1.27 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.28, and the impact drives the block a distance of 16.0 m before it comes to rest, what was the muzzle speed of the bullet in meters/second?
1 answer
1 answer