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?
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...
(The entire section contains 87 words.)
check Approved by eNotes Editorial
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.