a 2.0 kg block is released from rest at the top of a 22° inclined plane of height of 0.65 m. At the bottom of the plane it collides & sticks to a block of 3.5 kg. the two blocks slide a together...

Question

a 2.0 kg block is released from rest at the top of a 22° inclined plane of height of 0.65 m. At the bottom of the plane it collides & sticks to a block of 3.5 kg. the two blocks together slide a distance of 0.57 m across a frictionless horizontal plane in 0.5 seconds

1. find the speed of the 2.0 kg block at bottom of the inclined plane before it hits the other block
2. find the acceleration of 2.0 kg block while it is sliding down the inclined plane
3. find the coefficient friction

Answer 1

You have to work the problem backwards.

1) the two blocks go v = d/t with no friction and constant v (Given d and t)

2)There is conservation of momentum so m₁v₁ = (m₁ + m₂)v₁₂ (You know m's and v₁₂ from problem 1 solve for v₁ - answer #1)

3) Assuming constant acceleration you know that 2ad = v₁² d = h/sinθ from the geometry and you know v₁ - solve for a = answer #2)

4) I don't know if you are doing energy yet, here are 2 ways:

Energy mgh - F_fd = 1/2 m v₁² Energy initial - friction work = final energy (all kinetic) You can solve for the force of friction

Then the coefficient is solved for using F_f = μmgcosθ

or Forces: mgsinθ - mgμcosθ = ma The m cancels and you know everything but mu.

Please consider a tutor.