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14 December 2015

Cart on an incline: what qualifies as an "external force?"

When teaching about energy for AP Physics 1, one of the trickiest bits is defining an appropriate system, and then applying the work-energy theorem correctly to that system.  The question:

Hi Greg. From my understanding, an external force for a cart going down a [frictionless] incline would be the normal force acting on the cart. 

The weight is an internal or conservative force, so none of the external forces on a frictionless incline do work? I still consider Fg parallel to be an internal force for the system. Is this a correct assumption?

Not sure... gotta define your system first.

If your system is just the cart, then two external forces act: the weight (i.e. force of the earth on the cart), and the normal force.  Both are "external" forces because the forces are applied by objects that are not part of the defined system.  The normal force is perpendicular to displacement, so does no work.  The weight does work, because mg is parallel to the vertical component of displacement. This work is mgh, where h is the vertical component of displacement.  The cart acquires kinetic energy by the work-energy theorem -- the work done by the earth is equal to the cart's change in kinetic energy.

However -- if your system is the earth and cart together, then the only external force is the normal force, which does no work because it's perpendicular to displacement.  The work done by the earth on the cart is internal to the system, and conservative; so the system potential energy (equal to mgh) changes.  The system acquires kinetic energy by reducing potential energy, without any work done by external forces to change the total mechanical energy.

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