Question

A 1.0 kg wood block is launched up a wooden ramp that is inclined at a 40° angle. The block's initial speed is 11 m/s. (Use µk = 0.20 for the coefficient of kinetic friction for wood on wood.)

(a) What vertical height does the block reach above its starting point?
(b) What speed does it have when it slides back down to its starting point?

Answer 1

The kinetics formula that applies to this problem is
1. To find height, we use 1/2mv12−mg∗μ∗h sin θ = mgh where μ is equal to 0.20, v is 11 m/s and θ is 40 degrees. we cancel mass, h is equal to 5.47 meters.
2. The final speed is from 2(ug) * h sin θ = vf2
2(0.2*9.8) * sin 40 * 5.47= vf2 ; vf is equal to 3.71 m/s.