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# Work/energy problem with friction

A conservation of energy problem where all of the energy is not conserved. Created by Sal Khan.

## Want to join the conversation?

• •   No. Friction always opposes relative slipping between two objects. If one body is placed over another and both the bodies move with the same velocity, there is no slipping between them and friction does not act. Now if the lower body is pulled with some force and there is no friction between the ground and the lower body, the frictional force between the 2 bodies will actually push the upper body forward so that there is no relative slipping between the 2 bodies. Here friction is acting along the direction of motion.
However if there is friction between the ground and the lower body, it will depend whether the force applied is less than the max. frictional force between the ground and the body. If the applied force is more that the max. frictional force, the frictional force between the upper body and the lower body comes into play else there is no friction between the upper and lower body as there is no relative slipping between them.
• At ,How Is Initial Energy + Work done by friction = Final energy •  The system is losing energy to friction, and work is energy entering or exiting the system. Therefore, we will be adding all the work done on the system to the initial energy to achieve the final energy. In this problem, work done by friction will be negative because the force is applied in the opposite direction as the motion, and will decrease the initial energy, giving us a Final Energy result that is less than the initial energy because energy is lost to friction.
• When we are trying to find the work due to friction why are we using W=FD and not W=FDcos5? When is it appropriate to use either formula? • It's not W = FDcos5, it's W = FDcos(180). This is a common misconception, remember that the angle theta in W = FDcos(theta), is equal to the angle BETWEEN the force vector and the distance vector. Since the distance vector is facing the complete opposite direction of the force vector, then the angle between them will be 180 degrees. COS(180) evaluates to "-1", which is why Sal always says that the work done by friction is NEGATIVE.
• Is work negative when the direction of force is opposite to the direction of motion? If yes, then why do we always have a positive value for work done against resistance(which is opposite to the direction of motion)? • Work can be positive or negative but it is by convention. If you are looking at some system you often say work is positive if the work is done on the system or negative if the system is doing the work.

Also there is nothing about a scalar value that keeps it from being negative.
• What is a conservative force? How is it different from a non conservative force? Mathematically too? • What Andrew has written is correct, but I thought I'd just add something.

Let F be a force field (vector field) in three dimensions. One can say that F is a conservative force if and only if there exists a potential function ƒ for which grad(ƒ) = ∇ƒ = F. We can also say that F is a conservative force if and only if curl(F) = ∇ x F = 0.

If the force is conservative, as Andrew has pointed out, the path of the object is irrelevant. When dealing with conservative forces, what is more interesting is the initial and the final position of the object in question. This is useful when one for instance want to calculate the work done on an object moving from point A to point B through the force field. W = ∫ F · d r along C, where C is the path of the object from A to B. Instead of calculating this line integral, one can for instance find the function ƒ and evaluate ƒ(B) - ƒ(A).
• What about the length of the ramp? Why only the height? The length can easily be much longer? And the height still the same. Doesn't that affect the potential energy? • i only wanted the whole topic on friction
but i cant find the topic friction can anybody help me
i want the definition of friction • Why do you use W=Fd to calculate the final energy and not use W=Fcos theta x d when it's on an inclined ramp? I thought you include the angle if there is one and in this case it was 5 degrees. •  