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

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melehat | eNotes Newbie

Posted October 26, 2013 at 5:39 PM via web

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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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kavya--kammana | Student, Grade 12 | (Level 1) Valedictorian

Posted October 28, 2013 at 7:02 AM (Answer #2)

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

                                    500 x 9.8 x 50

                                  = 245000 J

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 26, 2013 at 5:48 PM (Answer #1)

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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 stored in the form of gravitational potential energy. The difference in the potential energy of the 500 kg elevator at the point it is initially and at the point 50 m upwards is m*g*h = 500*9.8*50 = 245000

The net work done on the elevator as it is moved upwards 50 m due to the force acting on it is equal to 245 kJ.

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tonys538 | Student, Undergraduate | TA | (Level 1) Valedictorian

Posted January 25, 2015 at 1:23 PM (Answer #3)

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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.

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