Multiply the force by the distance. The mass is irrelevant for this problem.
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.
Work = force x distance = Newtons x meters = 1937 Joules.
The work done is given by the formula Work = Force x Distance. Rearranging this formula we find Distance = Work / Force. Plugging in the values given, we get Distance = 68 joules / 4 newtons = 17 meters. So, the crate was moved 17 meters.
The reaction force is 100 newtons exerted by the crate on the boy in the opposite direction (Newton's third law).
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).
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.
Work = force x distance = Newtons x meters = 1937 Joules.
400 J
Force x distance = 100 x 2 = 200 newton-meters = 200 joules.
The work done is given by the formula Work = Force x Distance. Rearranging this formula we find Distance = Work / Force. Plugging in the values given, we get Distance = 68 joules / 4 newtons = 17 meters. So, the crate was moved 17 meters.
The reaction force is 100 newtons exerted by the crate on the boy in the opposite direction (Newton's third law).
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).
By using a pulley, Erin can reduce the amount of force needed to lift the crate. If Erin uses a single fixed pulley, she would still need to apply 50 newtons of force to lift the crate, but it would require less physical effort due to the change in direction of the force. If she uses a more complex pulley system, like a block and tackle, she could further reduce the amount of force needed, making it even easier to lift the crate.
yes
Fnet = F1+F2Force of man f1 = 10nForce of goat f2 = 10nSo,Fnet = 10+10Fnet=20n================================-- If Fred and the man are both pushing in the same direction, thenthe net force is 20 newtons.-- If they're not both pushing in the same direction, then the net forceis less than 20 N, but we don't have enough information to be able tonail down either the net force or its direction.
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.
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.