A factory worker moves a 26.0kg crate a distance of 5.20malong a level floor at

Question

A factory worker moves a 26.0kg crate a distance of 5.20malong a level floor at constant velocity by pushing horizontally onit. The coefficient of kinetic friction between the crate and thefloor is 0.300. What magnitude of force must the worker apply? How much work is done on the crate by the worker's push? How much work is done on the crate by friction? How much work is done by the normal force? How much work is done by gravity? What is the net work done on the crate? A factory worker moves a 26.0kg crate a distance of 5.20malong a level floor at constant velocity by pushing horizontally onit. The coefficient of kinetic friction between the crate and thefloor is 0.300. What magnitude of force must the worker apply? How much work is done on the crate by the worker's push? How much work is done on the crate by friction? How much work is done by the normal force? How much work is done by gravity? What is the net work done on the crate?

Explanation / Answer

= 0.30  = Force of friction/NormalForce Normal Force = 25 kg x 9.81 m/s2 = 255 N Force of Friction which the worker must overcome in orderto move the box = 0.30 x 255 N = 76.5 N Work done by the worker's push = F x d = 76.5 N x 5.20 m = 398 Joules Work done by frction = F x d = -398 Joules, the same as thepush as the worker in the opposing, since the crate is moving atconstant velocity. It's negative since the force of frictionis opposing the worker's push, it acts in the oppositedirection. Work done by the normal force and by gravity is zero, sincethe box is moving horizontally. The net work done on the crate is zero Newtons.

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