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A 500-kg elevator is pulled upward with a constant force of 5500N for a distance of 50.0 m. What is the net work done on the elevator?

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A 500 kg elevator is pulled upward with a constant force of 5500 N for a distance of 50.0 m

The net work done on the elevator as it moves upwards is...

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tonys538 | Student

The elevator with mass 500-kg is pulled upward with a constant force of 5500 N for a distance of 50.0 m.

Now the work done when a force F causes a displacement d, is given by W = F*d*sin x, where x is the angle between the vectors representing force and displacement.

Here, using the same formula and assuming displacement and force applied are in the same direction gives the work done as W = 5500*50 = 275 kJ.

But the increase in height of the elevator is 50 m. For this to happen, the work to be done on the elevator is equal to m*g*h = 500*9.8*50 = 245 kJ. This shows that the force acting on the elevator was not one that was acting vertically upwards.

kavya--kammana | Student

work done to lift the body= massx9.8xdistance ( F.S)

                                    500 x 9.8 x 50

                                  = 245000 J