PhysicsLAB: Work and Energy

According to Newton's Third Law, a force can be defined as the interaction between two objects: when object A exerts a force on object B, then object B exerts an equal but opposite force on object A. These forces are measured in a unit known as a newton.

The definition of this unit is based on Newton's Second Law which states that the acceleration an object experiences is directly proportional to the net force (sum of the forces) acting on the object and is inversely proportional to the object's mass; that is,

net F = ma

Refer to the following information for the next three questions.

Suppose a 5-kg mass, initially at rest, is accelerated to a final velocity of 8 m/sec in 4 seconds across a frictionless surface.

	What acceleration does it experience?

	How large a force was acting on the object to accomplish this task?

	Suppose the mass is now pulled across a second surface which is no longer smooth. On this rough surface the object encounters a constant frictional "retarding force" of 7 newtons as it is moves forward. How large would the new pulling force have to be to achieve the same acceleration?

When a force is exerted on an object and moves it through a distance we say that work is done on the object.

	How much work is done on our 5-kg mass when the object is pulled 16 meters across the original frictionless surface by a 10-N force?

	How much work is done on our 5-kg mass when the object was pulled 16 meters across the rough surface by a 17-N force?

	How much work is done on our 5-kg mass by the 7-N frictional force when the object was pulled 16 meters across the rough surface?

	If friction is said to do "negative work" because it decreases an object's velocity (that is, it opposes an object's forward motion), then how much "net work" was done in pulling the 5-kg mass 16 meters across the rough surface?

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy. That is, when an external force moves an object through a distance it does work on the object which is evidenced by a change in its velocity or its kinetic energy, KE = ½mv².

W_done = deltaKE

Recall from our lesson on mechanical energy that energy is measured in a unit called a joule which equals a kg m²/sec². This same unit is used to measure the work done on an object when a force moves it through a distance: nt m = (kg m/sec²) m = kg m²/sec².

	In our example, how much kinetic energy did the 5-kg mass gain as it was accelerated from 0 to 8 m/sec in 4 seconds?

	Could this acceleration have occurred in 16 meters?

Another graph of F vs d- questions about work, change in KE, final velocity, acceleration

If an object is moving at a constant velocity then there must be at least two forces acting on it - one force causing a "forward" acceleration and another force causing an "opposing" acceleration. These forces must balance to a net force of zero resulting in a corresponding acceleration of zero. Note that each force is doing work on the object but their "work" cancels to no net change in the object's kinetic energy.

When a path-independent force, or conservative force (for example, the pull of gravity), acts on an object, the change in the object's kinetic energy is related to a change in its potential energy.

PE_lost = KE_gained