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How much power is required to raise a 30 kg crate a vertical distance of 6 m in a time of 4 seconds?
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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 work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
A forklift picks up a crate using 400 N It moves the crate 20 meters in 50 seconds How much power is used?
From the question, it's hard to tell whether the 20 meters is the vertical lift, or a horizontal transfer that occurs after the lift.If the 20 meters is the vertical lift (performed by a very large fork-lift in a shop with a very high ceiling):Energy = work = 400 N times 20 m = 8,000 Newton-meters = 8,000 joules8,000 joules in 50 seconds = 8,000 / 50 = 160 joules per second = 160 watts = about 0.214 horsepower.If the 20 meters is a horizontal ride after the lift is complete, then that part of the move consumes nominally no energy or power. No force is required to move an object perpendicular to the force of gravity. Whatever force is applied initially, to get the crate moving, is returned at the end of the 20 meters, when reverse force must be applied to the crate in order to make it stop moving.
If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
Impossible:)
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.
What is the power rating of a machine to lift a crate in 5 seconds?
That really depends on the weight of the crate. Also, on how high you want to lift it. Calculate the energy required to lift the crate with the formula for gravitational potential energy: PE = mgh (mass x gravity x height) Then divide this by the 5 seconds to get the minimum power required. (The actual power is somewhat larger, for various reasons - the initial acceleration required, and losses due to friction.)
What is the potential energy of a crate lifted with a force of 100 newtons a vertical distance of 2 meters?
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.
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.
If a 50 crate was lifted to a height of 10 meters and it took 5 seconds to lift the crate using a machine what was the power rating of the machine?
Please use the formula for gravitational potential energy (PE = mgh) to calculate the energy required. Then divide that by the time to get the power.
How much work is done in lifting a 50kg crate a vertical distance of 10 meters?
Potential Energy is given by the fourmulai PE=MGH where M=mass in kilos, G=the force of gravity in Netwons (9.8N) and H=height in meters. So 50*9.8*10=4900joules. A Watt is a unit of power that =1 joule per second. So 4900 joules divided by 5 seconds = 980 Watts, not allowing for losses due to friction etc.
A forklift picks up a crate using 400 N It moves the crate 20 meters in 50 seconds How much power is used?
From the question, it's hard to tell whether the 20 meters is the vertical lift, or a horizontal transfer that occurs after the lift.If the 20 meters is the vertical lift (performed by a very large fork-lift in a shop with a very high ceiling):Energy = work = 400 N times 20 m = 8,000 Newton-meters = 8,000 joules8,000 joules in 50 seconds = 8,000 / 50 = 160 joules per second = 160 watts = about 0.214 horsepower.If the 20 meters is a horizontal ride after the lift is complete, then that part of the move consumes nominally no energy or power. No force is required to move an object perpendicular to the force of gravity. Whatever force is applied initially, to get the crate moving, is returned at the end of the 20 meters, when reverse force must be applied to the crate in order to make it stop moving.
If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
Impossible:)
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.
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
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 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.
A crate is at rest on the ground what force or forces are acting on the crate?
The weight of the crate is acting downward on the ground and the ground is exerting a force equal to the weight of the crate upward on the crate.
