A 280-kg load is lifted 24.0 m vertically with an acceleration a=0.120 g by a si

Question

A 280-kg load is lifted 24.0 m vertically with an acceleration a=0.120 g by a single cable.

Part A:Determine the tension in the cable.

Express your answer to three significant figures and include the appropriate units.

Part B: Determine the net work done on the load.

Express your answer to three significant figures and include the appropriate units.

Part C:Determine the work done by the cable on the load.

Express your answer to three significant figures and include the appropriate units.

Part D:Determine the work done by gravity on the load.

Express your answer to three significant figures and include the appropriate units.

Part 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.

Explanation / Answer

1) tension in the cable

ma = T -mg

T = m(a+g) = 280*(0.12*9.8+9.8) = 3073.28 N

2) work done = F*d = 3073.28*24 = 73758.72 J

3) work done by the cable also same but negative sign ie -73758.72 J

4) Wg = 280*9.8*24 = 65856 J

5) v^2-U^2 =2as

here u = 0

v = sqrt(2as) = sqrt(2*0.12*9.8*24) = 7.5 m/sec

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