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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).
A crate is pulled across a horizontal floor by a rope At the same time the crate pulls back on the rope in accord with Newton's 3rd law Does the work done on the crate by the rope then equal zero?
No, the work done on the crate by the rope is not zero. The work done is equal to the force exerted by the rope multiplied by the distance the crate is pulled. The fact that the crate pulls back on the rope in accordance with Newton's Third Law does not cancel out the work done by the rope.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
How much power is required to raise a 30 kg crate a vertical distance of 6 m in a time of 4 seconds?
The work done in lifting the crate is equal to its change in potential energy: ( \text{Work} = \text{Force} \times \text{distance} = m \cdot g \cdot h ). The power required is the work done divided by the time taken: ( \text{Power} = \frac{\text{Work}}{\text{time}} ). Plug in the values to calculate the power required.
How much work does a crane do as it lifts a 240 kilogram crate from ground level to a height of 165 meters?
The work done by a crane lifting a 240 kg crate to a height of 165 meters is equal to the change in potential energy of the crate. The work done can be calculated using the formula: Work = force x distance. In this case, the force is equal to the weight of the crate (240 kg x 9.8 m/s^2) and the distance is 165 meters.
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).
How much work is done on a 100 kg crate that is hoisted 3 m in a time of 2 s?
w=2.9kilojoules..
A crate is pulled across a horizontal floor by a rope At the same time the crate pulls back on the rope in accord with Newton's 3rd law Does the work done on the crate by the rope then equal zero?
No, the work done on the crate by the rope is not zero. The work done is equal to the force exerted by the rope multiplied by the distance the crate is pulled. The fact that the crate pulls back on the rope in accordance with Newton's Third Law does not cancel out the work done by the rope.
If it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
We have no way of knowing what power the machine was rated for, but with the information given in the question, we can calculate the power it delivered during the crate-lift: It was (1.96) x (mass of the crate in kilograms) x (distance the crate was lifted in meters) watts.
How much power is required to raise a 30 kg crate a vertical distance of 6 m in a time of 4 seconds?
The work done in lifting the crate is equal to its change in potential energy: ( \text{Work} = \text{Force} \times \text{distance} = m \cdot g \cdot h ). The power required is the work done divided by the time taken: ( \text{Power} = \frac{\text{Work}}{\text{time}} ). Plug in the values to calculate the power required.
How much work does a crane do as it lifts a 240 kilogram crate from ground level to a height of 165 meters?
The work done by a crane lifting a 240 kg crate to a height of 165 meters is equal to the change in potential energy of the crate. The work done can be calculated using the formula: Work = force x distance. In this case, the force is equal to the weight of the crate (240 kg x 9.8 m/s^2) and the distance is 165 meters.
How much work does someone need to do if a crate moves 20 meters with force of 200 newtons?
The work done on the crate would be 4000 joules (W = F x d), as work is the product of force (200 N) and distance (20 m).
When you push a crate with N across a M floor how much work is done?
The work done when pushing a crate with a force N across a distance M on a floor depends on the angle between the force and the direction of motion. If the force is applied in the same direction as motion, work done is N * M. If the force is applied at an angle, work done is N * M * cos(theta), where theta is the angle between the force and direction of motion.
A student applies a 20 newton force to move a crate at a constant speed of 4 meters per second across a rough floor how much work is done by the student on the crate in 6 seconds what is the answer?
80 J
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 50kg crate is lifted to a height of 10 meters if it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
The work done to lift the crate is equal to the gravitational potential energy gained: Work = force x distance = weight x height. Here, Work = 50kg x 9.8m/s^2 x 10m = 4900 Joules. Power is work done per unit time, so Power = Work / time = 4900J / 5s = 980 Watts. Therefore, the power rating of the machine is 980 Watts.
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.
