A 285-kg load is lifted 21.0 m vertically with an acceleration a=0.150 g by a si

Question

A 285-kg load is lifted 21.0 m vertically with an acceleration a=0.150 g by a single cable A) Determine the tension in the cable. Express your answer to three significant figures and include the appropriate units. B) Determine the net work done on the load. Express your answer to three significant figures and include the appropriate units. Wnet = C) Determine the work done by the cable on the load. Express your answer to three significant figures and include the appropriate units. D) Determine the work done by gravity on the load. Express your answer to three significant figures and include the appropriate units. E) Determine the final speed of the load assuming it started from rest. Express your answer to three significant figures and include the appropriate units. v =

Explanation / Answer

a) Using newton's second law,

T - mg = ma

T = ma + mg

T = m(g+a)

T = 285 (9.8 + 1.47)

T = 3211.95 N

b) Wnet = Fnet*h

Wnet = (T-mg)*h

Wnet = (3211.95 - 2793) * 21

Wnet = 8797.95 joules

c) Work done by cable = T*h

Wcable = 3211.95*21

Wcable = 67450.95

d) work done by gravity = -mgh

Wgravity = - 285*9.8*21

Wgravity = - 58653 joules.

e) Using work energy theorem

W = change in energy

  8797.95 = 0.5*285*v2

v = 7.857 m/s

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