Question

A 285- kg load is lifted 24.0 m m vertically with an acceleration a =0.130 g g by a single cable.

Part A Determine the tension in the cable.

Determine the net work done on the load.

Determine the work done by the cable on the load.

Determine the work done by gravity on the load.

Determine the final speed of the load assuming it started from rest.

Answer 1

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A)

Fnet = ma

T - mg = ma

T = mg + ma

T = m ( g + a)

T = 285 ( g + 0.130g)

put g = 9.81 m/s²

so,

T = 3156 N

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B)

Net force = T - mg = 363.1 N

so,

W = 363.1 * 24

W = 8714.16 J

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c)

work done by cable

W = T * d

W = 3156 * 24

W = 75744 J

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d)

work done by gravity

W = 285 * 9.8 * 24 * cos 180

W = - 67032 J

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e)

net work done = change in K.E

8714.16 = 1/2 * 285 * v²

so,

v = 7.82 m/s