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Zackery Schumm

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Q: How much work is performed when a 60 kg crate is pushed with 20 m with a force of 20 N?
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How much work is performed when a 60 kg crate is pushed 20 m with a force of 40 N?

400 J


How much work is performed when a 60 kg crate is pushed 15 m with a force of 30 N?

Multiply the force by the distance. The mass is irrelevant for this problem.


How much work is performed when 100 kg crate is pushed 30 meters with a force of 40 newtons?

Multiply the force by the distance. The mass is irrelevant for this problem.


How much work is performed when a 25-kilogram crate is pushed 5 meters with a force of 15 newtons?

Work = (force) x (distance) = (15) x (5) = 75 joules.The mass of the crate is irrelevant.


How much work is performed when a 50 kg is pushed 15 m with a force of 20 N?

If a force of 20 N acts through a distance of 15 m, then it does (20 x 15) = 300 joules of work.The other facts of the case, such as the mass of the crate being pushed, don't matter.


If you apply a force of 220 N to the lever how much force is applied to lift the crate?

If the perpendicular distance from the point of application of the force to the fulcrum is x metres and the perpendicular distance from the crate to the fulcrum is y metres, then the force applied on the crate is 220*x/y N.


If a worker pushes horizontally on a large crate with a force of 298 newtons and the crate moved 6.5 meters how much work do he do?

Work = force x distance = Newtons x meters = 1937 Joules.


How much work is done when a 100N force pushes a crate 5m across a factory floor?

Work performed = Force x displacement = 100 x 5 = 500 J (joule)


How much work in joules is performed when a force of 800.0 N is exerted while pushing a crate across a level floor for a distance of 1.5 meters?

We must assume that the force pushes parallel to the floor.Work = (force) x (distance) = (800) x (1.5) = 1,200 newton-meters = 1,200 joules


What is comperession?

compression is the force pushed on and object. The formula for how much pressure during compression is: force/area.


Pull horizontally on a crate with a force of 140 N and it slides across the floor in a dynamic equilibrium How much friction is acting on the crate?

If the crate isn't accelerating ... i.e. sliding at a constant speed, not speeding up or slowing down ...then the forces on it are balanced. The pseudo-force of friction is 140N in the direction opposite toits speed.


How much work is performed when a 50kg is pushed 15m with a force of 20n?

That depends also on the direction of force, though it simplifies greatly if force is all the way applied exclusively in the direction of movement. Work doesn't depend on the mass too. Formula is(assuming the direction of applied force is the same as of displacement): W = F * r, where W - work, F - force, r - displacement. For given data, it will be: W = 20 * 15 = 300 J