suppose you shoot a bullet of mass 2g into a wooden block of mass 0.498 kg resting on the table. the bullet gets stuck in the wood, and the block slides a distance of 0.2m before it comes to rest again. how much work is done by the friction if the block with the bullet slides a distance of 0.2m on te table with a frictional coefficient of 0.4?
2007-12-26 01:57:29 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hmmm ...a peoblem with redundant conditions?
1. Conservation of momentum
m1V1= V(m1 + m2)
however we do not know initial velocity of the bullet.
2. Kinetic energy transfered to work done W against friction.
The force of friction f is
f= uN= u mg
u - coefficient of friction
m - total mass
g acceleration due to gravity
We have work W as product of force of friction f times distance s,
W= f s = u (m1 + m2) g s
W= 0.4 ( . 002 + .498) x 9.81 x 0.2=0.392 Joules
2007-12-26 02:13:24 · answer #1 · answered by Edward 7 · 4⤊ 0⤋