A 3.50 kg block is pulled along a moving conveyor belt at a constant speed of 0.500 m/s relative to a stationary observer while the belt moves at a constant speed of 0.200 m/s in the opposite direction. If the coefficient of kinetic friction is 0.400, what is the magnitude of the mechanical energy dissipated, in J, caused by the force of friction on the block in 8.00 s?

Question

Physics

Duncan Chaney

7 July, 05:19

A 3.50 kg block is pulled along a moving conveyor belt at a constant speed of 0.500 m/s relative to a stationary observer while the belt moves at a constant speed of 0.200 m/s in the opposite direction. If the coefficient of kinetic friction is 0.400, what is the magnitude of the mechanical energy dissipated, in J, caused by the force of friction on the block in 8.00 s?

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Answers (1)

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Lucian Rivera · Answer 1

76.8 J

Explanation:

m = 3.50 kg

velocity of block relative to stationary observer = 0.5 m / s

velocity of belt = 0.2 m / s opposite direction

velocity of block with respect to belt = 0.5 + 0.2 = 0.7 m / s

time, t = 8 s

distance moved = velocity x time

d = 0.7 x 8 = 5.6 m

coefficient of friction, μ = 0.

Work done by friction force

W = μ x mg x d = 0.4 x 3.5 x 9.8 x 5.6 = 76.832 J

Thus, the magnitude of mechanical energy dissipated is equal to the work done by friction force = 76.8 J