# Work as Change in Kinetic Energy vs Change in PE

1. Sep 16, 2014

### Hereformore

Net Work = Change in Kinetic Energy. What about potential energy?

1. The problem statement, all variables and given/known data

When we say net work = change in kinetic energy, does that mean that if the height of an object changes (and its initial and final velocity) do not change, then the work done also equals the potential energy?

Sometimes I see texts saying work = change in potential energy, but i thought it was change in kinetic energy. Do they mean change in potential energy = work done by the force(s) that change the height of the object and change in kinetic energy = NET Work done?

2. Relevant equations
U = mgh
KE = 1/2mv^2

Work = Change in KE OR Change in PE (?)

3. The attempt at a solution

What confuses me is that if i raise a book from a height of 0 m to 5m, then the potential energy definitely changes but the kinetic energy does not change (0m/s initially and 0 m/s final).

So Change in kinetic energy in this case cannot equal change in potential. So here work does NOT equal change in kinetic energy.

Last edited: Sep 16, 2014
2. Sep 17, 2014

### ehild

When you speak about work, you need to specify the force that does the work and the object the work is done on.
The work-energy theorem states that the change of the kinetic energy of a point mass (particle) is equal to the work of all the forces acting on it. If a vertical upward force F acts on a particle and it rises from h1 to h2 , the work of the applied force is Wa= F(h2-h1) and the work of gravity is Wg = -mg(h2-h1). The work of all the forces is equal to the change of the KE:

KE2-KE1=Wa+Wg=(F-mg)(h2-h1).

If the KE does not change, the net work is zero, so Wa+Wg=0, the applied force is negative of the force of gravity.

We define the potential with respect to a place where the PE is set equal to zero. If it is the ground, the PE of a body at a point A is the work done by the gravitational force while the body moves from A to the ground. It is PE(A)=mghA.

If the body moves from h1 to h2, the potential energy changes by ΔPE=PE(h2)-PE(h1)=mg(h2-h1). That is equal the negative of the work done by gravity: Wg=-ΔPE. You can write the Work-Energy theorem in the form
Wa-ΔPE=ΔKE, that is, Wa=ΔPE+ΔKE. The work of the applied force is equal to the change of the total mechanical energy. In case the KE does not change, the work done by the applied force is equal to the change of the potential energy.

ehild

Last edited: Sep 17, 2014
3. Sep 17, 2014

### Simon Bridge

I'd like to add slightly to what ehild has said, at the risk of being a bit confusing:
... that action does involve a change in KE ... between 0m and 5m the book must have some non-zero velocity. It follows that you did some work in addition to that against gravity by accelerating and decelerating the book.

It is possible to lift a book through a height without changing KE, and doing work, though.
This requires the book to already be in motion at a constant speed...

4. Sep 17, 2014

### CWatters

What Simon said. A human lifting a book is complicated by the fact that a human cant recover the ke during the deceleration phase. In many other systems that energy can be recovered. Humans are also inefficient and consume energy even when not doing work lifting books. Sometimes best not to use humans in examples because they are so non ideal machines.