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How much work did the movers do horizontally pushing a 150 kg crate 10.4 m across a rough floor without acceleration if the effective coefficient of friction was 0.70?
The work done by the movers can be calculated using the work-energy principle. The work done can be found by multiplying the force of friction by the distance the crate was moved. The force of friction is the product of the coefficient of friction and the normal force (weight of the crate). The work done will be equal to the force of friction multiplied by the distance moved.
A student pushed a box 32.0 m across a smooth horizontal floor using a constant force of 124 N. If the force was applied for 8.00 s how much power was developed to the nearest watt?
First you need to calculate the work required. This is simply force x distance in this case.Once you have that, divide the work by the time.
How much work is done on a 40 N rock that won't move when pushed?
Work = (force) x (distance)Without motion, there is no work, no matter how great the force.
Infer use newtons second law to determine how much force is being applied to an object that is traveling at a constant velocity?
If there is no force against motion,applied force is zero. If there is force against motion,applied force is equal and opposite to that force.
How do you measure force in safety shoe?
get a force meter The most common tests done on safety shoes are: - impact resistance - compression resistance - penetration resistance For impact resistance a weight is dropped on the the toe-cap. Then it's measured how much millimeter the toe cap was indented. For compression resistance a gradual force is applied on the toe cap until it breaks or a certain number of millimeters of indent is reached For penetration resistance a nail is pushed through the sole and the protective mid plate, again the indent and force applied is measured.
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 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.
How much work is performed when a 60 kg crate is pushed with 20 m with a force of 20 N?
400 J
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).
