1. A 4.0 kg block is placed on an inclined plane ofinclination 60 o at a height
ID: 1682147 • Letter: 1
Question
1. A 4.0 kg block is placed on an inclined plane ofinclination 60o at a height 2.0 above the ground. Theblock starts from rest and slides down the plane. If thecoefficient of kinetic friciton is 0.2, what is the frictionalforce acting on the block?2. the work done by the frictional force on theblock?
3. the kinetic energy of the block at the bottom of theplane?
4. the speed of the block at the bottom of the inclinedplane?
2. the work done by the frictional force on theblock? 3. the kinetic energy of the block at the bottom of theplane?
4. the speed of the block at the bottom of the inclinedplane?
Explanation / Answer
mass m = 4 kg angle = 60 degrees height h = 2 m distance along the incline S = h / sin = 2.309 m initila velocity u = 0 coeffcient of kinetic friction = 0.2 frictional force f = mg cos = 3.92 N (2).the work done by the frictional force on the block W= fS = 9.05 J 3. the kinetic energy of the block at the bottom of theplane K = Initial potential energy - W = mgh - 9.05 = 78.4 J - 9.05 J = 69.35 J4. the speed of the block at the bottom of the inclinedplane v = [ 2K / m] = 5.888 m / s
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