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A mover’s dolly is used to transport a refrigerator up a ramp into a house. The refrigerator has a mass of 115 kg. The ramp is 2.10 m long and rises 0.850 m. The mover pulls the dolly with a force of 496 N up the ramp. The dolly and ramp constitute a machine. a. What work does the mover do? b. What is the work done on the refrigerator by the machine? c. What is the efficiency of the machine?

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User Nao
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The mover exerts a force of 496 N on the dolly to transport the 115 kg refrigerator up the 2.10 m ramp that rises 0.850 m. To determine the work done by the mover, the formula W = Fd cos(theta) is used, where W is work, F is force, d is distance, and theta is the angle between the force and displacement vectors. In this case, the work done by the mover is W = (496 N)(2.10 m) cos(30 degrees) = 820 J.

The work done on the refrigerator by the machine can be found using the formula W = Fd. The force used to move the refrigerator up the ramp is the weight of the refrigerator, which is F = mg = (115 kg)(9.81 m/s^2) = 1127 N. The distance the refrigerator moves up the ramp is the vertical displacement, which is 0.850 m. Therefore, the work done on the refrigerator is W = (1127 N)(0.850 m) = 959 J.

To find the efficiency of the machine, the formula efficiency = (work output / work input) x 100% is used. In this case, the work output is the work done on the refrigerator, which is 959 J, and the work input is the work done by the mover, which is 820 J. Therefore, the efficiency of the machine is (959 J / 820 J) x 100% = 117%. This value is greater than 100% because some of the work input is lost as friction and other factors.
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User David Watts
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