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Crank
3 years ago
13

How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if t

he effective coefficient of friction was 0.50
Physics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

7.1 J

Explanation:

From the question,

Work done by the mover  = work done in pushing the crate + work done against friction

W = W'+Wf................. Equation 1

W = mgd+mgμd............ Equation 2

W = mgd(1+μ)................ Equation 3

Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of  friction.

Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5

constant: g = 9.8 m/s²

Substitute these values into equation 3

W = 46×9.8×0.0105(1+0.5)

W = 7.1 J

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Explanation:

Given:

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<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

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As we know that the frictional force acting on the body is always in the opposite direction:

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