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## Understanding Work Done

 Quote by Nubcake I was more concerned about the direction and the force ; doesn't the weight of the barrel act in the opposite direction of movement?
Yes. So the work done by gravity is negative (-360 J). By definition, the change in gravitational potential energy is equal to the negative of the work done by gravity. So, the change in potential energy is +360 J (potential energy increases when gravity does negative work).

The work done by the person pushing the thing up the ramp is positive, since the displacement and force are in the same direction. Let's assume that there is negligible friction. Let's also assume that the object is pushed up the ramp at a constant speed (neglecting the initial acceleration to get it moving). Then the forces parallel to the ramp have to be balanced, which means that the person pushes with a force equal to mgsinθ, which is the component of the weight that acts "down the ramp." The work done is then mgsinθ*d, where d is the distance travelled along the ramp. But from basic trigonometry, d = h/sinθ, so W = mgsinθ*(h/sinθ) = mgh.

The person does an amount of work equal to mgh = +360 J (but the ramp allows him to do so with a smaller force than if he just lifted it vertically).

Notice that the person does +360 J of work, and gravity does -360 J of work on the object, so the NET work done on the object is 0 (which makes sense because it has 0 net force on it). This explains why it does not gain any kinetic energy. (Recall that the work-energy theorem says that the work done on an object is equal to its change in kinetic energy).