A 6.0 kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline, it's speed is vf. At what distance from the top of the incline is the speed of the block 0.5vf?
Since Vf^2 = 2*a*S Given S=3.6m, thus a = Vf^2/(2*3.6) a = Vf^2/7.2
Let d be the distance along the slope at which the velocity is 0.5Vf, then (0.5Vf)^2 = 2*a*d or d = (0.5*Vf)^2/(2*a) with a = Vf^2/7.2, we have d = 0.9 m
