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A person pulls a 2.6-kg crate 36 m along a horizontal floor by a constant force Fp= 87 N which acts at a 38° angle as shown in Fig. 6-3. The floor is rough and exerts a friction force FFR = 51 N

Determine the total work done by combining each force acting on the crate

User Subodh
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1 Answer

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The total work done on the crate can be determined by calculating the work done by each force separately and then adding them together.

1. Work done by the applied force Fp = (Fp * cos(theta)) * d

where Fp = 87 N, theta = 38 degrees and d = 36 m

cos(theta) = cos(38) = 0.81915

Work done by Fp = (87 N * 0.81915) * 36 m = 2789.5 J

2. Work done by the friction force FFR = -(FFR * d)

where FFR = 51 N and d = 36 m

Work done by FFR = -(51 N * 36 m) = -1836 J

3. Work done by gravity

Work done by gravity = forcedistancecos(theta)

force of gravity = mg

where m = 2.6 kg and g = 9.8 m/s^2

Work done by gravity = 2.69.836cos(90) = 0 J

4. The total work done is the sum of the work done by each force:

Total work done = work done by Fp + work done by FFR + work done by gravity

Total work done = 2789.5 J + (-1836 J) + 0 J = 953.5 J

The total work done on the crate is 953.5 J.

User MuraliMohan
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