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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.
yes
yes
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.
Work = force x distance = Newtons x meters = 1937 Joules.
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.
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.
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.)
yes
yes
80 J
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.
Work = force x distance = Newtons x meters = 1937 Joules.
Impossible:)
Work = (force) x (distance) = (15) x (5) = 75 joules.The mass of the crate is irrelevant.
420,000 j
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.