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How much work is performed when a 60 kg crate is pushed with 20 m with a force of 20 N?
400 J
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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?
The work done is calculated as the force applied multiplied by the distance traveled in the direction of the force. In this case, the work done is 75 Joules (15 N * 5 m) when pushing the 25 kg crate with a force of 15 Newtons over a distance of 5 meters.
If we push a crate at a constant velocity how do we know how much friction acts on the crate compared to our pushing force?
If the crate is moving at a constant velocity, then the force of friction acting on the crate is equal in magnitude and opposite in direction to the force you are applying to push the crate. This means that the force you apply to push the crate is balancing out the force of friction acting against it. By measuring the force you are exerting and observing the constant velocity of the crate, you can infer the magnitude of the friction force.
If you push a crate with a force of 100 N and it slides at a constant velocity how much is the friction acting on the crate?
If the crate is moving at a constant velocity, the friction force is equal in magnitude but opposite in direction to the pushing force, so it is also 100 N. This is because the two forces are balanced and there is no net force acting on the crate.
Once the crate is sliding how hard do you push to keep is moving at constant valocity?
You should push with a force equal to the force of friction acting on the crate. This will counteract the friction force and allow the crate to continue moving at a constant velocity. Pushing with a greater force will accelerate the crate, while pushing with a force lower than the frictional force will cause it to decelerate.
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 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 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.
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 a 25-kilogram crate is pushed 5 meters with a force of 15 newtons?
The work done is calculated as the force applied multiplied by the distance traveled in the direction of the force. In this case, the work done is 75 Joules (15 N * 5 m) when pushing the 25 kg crate with a force of 15 Newtons over a distance of 5 meters.
If we push a crate at a constant velocity how do we know how much friction acts on the crate compared to our pushing force?
If the crate is moving at a constant velocity, then the force of friction acting on the crate is equal in magnitude and opposite in direction to the force you are applying to push the crate. This means that the force you apply to push the crate is balancing out the force of friction acting against it. By measuring the force you are exerting and observing the constant velocity of the crate, you can infer the magnitude of the friction force.
If you push a crate with a force of 100 N and it slides at a constant velocity how much is the friction acting on the crate?
If the crate is moving at a constant velocity, the friction force is equal in magnitude but opposite in direction to the pushing force, so it is also 100 N. This is because the two forces are balanced and there is no net force acting on the crate.
Once the crate is sliding how hard do you push to keep is moving at constant valocity?
You should push with a force equal to the force of friction acting on the crate. This will counteract the friction force and allow the crate to continue moving at a constant velocity. Pushing with a greater force will accelerate the crate, while pushing with a force lower than the frictional force will cause it to decelerate.
You are pushing a kg wooden crate across the floor. the force of sliding friction on the crate is 90 N. how much force must you exert on the crate to keep it moving with a constant velocity?
To keep the crate moving with constant velocity, the force you exert must balance the force of sliding friction. In this case, you must exert a force of 90 N in the opposite direction of the sliding friction, so the net force on the crate is zero and it remains in motion at a constant velocity.
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.
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 done on a 20-N crate that you lift 2 meters?
The work done on the crate would be 40 joules (work = force x distance).
