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Unformatted text preview: 45. (a) The work done on the block by the force in the rope is, using Eq. 7-7,
W = Fd cos θ = (7.68 N)(4.06 m) cos15.0° = 30.1J. (b) Using f for the magnitude of the kinetic friction force, Eq. 8-29 reveals that the increase in thermal energy is ΔEth = fd = (7.42 N)(4.06 m) = 30.1J. (c) We can use Newton's second law of motion to obtain the frictional and normal forces, then use μk = f/FN to obtain the coefficient of friction. Place the x axis along the path of the block and the y axis normal to the floor. The x and the y component of Newton's second law are x: F cos θ – f = 0 y: FN + F sin θ – mg = 0, where m is the mass of the block, F is the force exerted by the rope, and θ is the angle between that force and the horizontal. The first equation gives
f = F cos θ = (7.68 N) cos15.0° = 7.42 N and the second gives
FN = mg – F sin θ = (3.57 kg)(9.8 m/s2) – (7.68 N) sin15.0° = 33.0 N. Thus, μk = f 7.42 N = = 0.225 . FN 33.0 N ...
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