How much work did the movers do (horizontally) pushing a 42.0- kg crate 10.2 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60?

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Gerry Eng · Answer 1 · 2023-09-11T10:11:45+0000

Answer:

2521.6 J of work

Step-by-step explanation:

Normal force = 42 kg * 9.81 m/s^2 = 412 N

Friction force = 412 * .6 = 247.2 N

Work = Force * distance = 247.2 * 10.2 M = 2521.6 J