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.
Work = (force) x (distance)
The minimum force required to lift the crate is equal to its weight.
Work = (weight) x (distance) = (mass x gravity) x (Height) = (50 x 9.807) x (10) = 4,903.5 joules
The crate's potential energy is 4,900 joules.
And what is the question?
And what is the question?
And what is the question?
And what is the question?
Work=F x d
Therefore; Work= 20 x 10 = 200 Joules
If that task is done on earth, then it requires 4,900 hundred joules of energy.
The crate's PE is 4900 joules.
9800 or 9,800
And what is the question?
735 J
Work = force x distance = Newtons x meters = 1937 Joules.
Work = (force) x (distance) = (15) x (5) = 75 joules.The mass of the crate is irrelevant.
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.
Multiply the force by the distance. The mass is irrelevant for this problem.
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.
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.
Work = force x distance = Newtons x meters = 1937 Joules.
Work = (force) x (distance) = (15) x (5) = 75 joules.The mass of the crate is irrelevant.
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.
You don't actually lift the crate. You drag it. To do so you touch the crate with your stylus and without lifting the stylus drag your Jonas to the spot you want to take the crate to and release.
Multiply the force by the distance. The mass is irrelevant for this problem.
lifting a large crate tied with rope into a ship
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 is equal to force x distance. If the force is specified in Newtons, and the distance in meters, then the work is in Joules.Work is equal to force x distance. If the force is specified in Newtons, and the distance in meters, then the work is in Joules.Work is equal to force x distance. If the force is specified in Newtons, and the distance in meters, then the work is in Joules.Work is equal to force x distance. If the force is specified in Newtons, and the distance in meters, then the work is in Joules.
We must assume that the force pushes parallel to the floor.Work = (force) x (distance) = (800) x (1.5) = 1,200 newton-meters = 1,200 joules
yes
yes