A block of mass m sliding down an incline at constant speed is initially at height h above the ground as shown in the figure. The coefficient of kinetic friction between the mass and the incline is µ. If the mass continues to side down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reached the bottom of the incline?
A. mgh/µ
B. mgh
C. µmgh/sin θ
D. mgh sin θ
E. 0
(GR9677 #06)
mg sin θ − Fr = ma
At a constant speed, a = 0
Fr = mg sin θ
Energy dissipated = Work done by the frictional force
W = Fr ⋅ s
s = length of the inclined surface = h/sin θ
W = mg sin θ (h/sin θ) = mgh
Answer: B
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