Potential energy

The gravitational potential energy Ug is defined as the negative of the work done by the gravitational force, or the work done by an applied force canceling the gravitational force, in displacing an object from a reference position.  If the gravitational force acting on an object is pointing in the -y direction with constant magnitude mg, and the reference position is at y = 0, then Ug = mgy.

The zero of the gravitational potential energy, i.e. the reference position, is chosen arbitrarily.  However a difference in gravitational potential energy

∆Ug = mg∆y = mg(yf - yi)

is uniquely defined.

The gravitational potential energy gained by an object being lifted is equal to the work done on the object by an applied force, which exactly cancels gravity.  The applied force does positive work.  When the object is being lifted, the gravitational force does negative work.  The gravitational potential energy can be converted back into other forms by letting the gravitational force do positive work.  If we let the object fall towards the ground it will lose potential energy and gain kinetic energy.

Problem:

A 100 N crate sits on the ground and is attached to one end of a rope.  A person on a balcony pulls up on the rope, lifting the crate a distance of 3 m.  What is the change in the gravitational potential energy of the crate?

Solution:
∆Ug = mg∆y= 100 N* 3m = 300 J.  (Note:  100 N is the weight of the crate, i.e. the gravitational force mg acting on it.  The mass of the crate is m = 100 N/9.8 m/s2 = 10.1 kg.)

Problem:

A 100 N crate sits on the ground and is attached to one end of a rope.  A person on a balcony pulls up on the rope with a constant force of 110 N lifting the crate a distance of 3 m.
What is the kinetic energy of the crate when it reaches the height of 3 m?

Solution:
The net force on the crate is the difference between upward force from the rope and the downward force of gravity.
Fnet = 110 N - 100 N = 10 N.   Wnet = 10N * 3 m = 30 J = ∆K.
Since the crate starts with zero kinetic energy, its final kinetic energy is 30 J.
The applied force does 110 N * 3 m = 330 J of work.  300 N is converted into potential energy and 30 N into kinetic energy.


Conservative Forces

The gravitational potential energy does not depend on the path.

Example:

imageWe do the same work lifting the motorcycle straight up onto the bed of the truck or rolling it up a ramp.

The gravitational force near the surface of the earth points vertically downward.  We can approximate any path between two points P1 and P2 by arbitrarily small vertical and horizontal segments.  The total horizontal and the total vertical displacement are the same for each path.  The gravitational force does no work along the horizontal segments of the path, since here it is perpendicular to displacement vector.  It does work Wi = -mg∆yi along each vertical segment of length ∆yi.  The total work done by the gravitational force is W = -mg∆ytotal = -mgh.  The change in potential energy is Ug = mgh.

A potential energy function is a function of the position of an object.  It can be defined only for conservative forces.  A force is conservative if the work it does on an object depends only on the initial and final position of the object and not on the path.  The gravitational force is a conservative force.  The potential energy function associated with the gravitational force near the surface of the earth is Ug = mgy if the reference point is chosen at y = 0.  (Ug only depends on the position of an object, not on how the object reached that position.)

Another example of a conservative force is the force exerted by a spring.  The elastic potential energy function is Us = ½kx2, where x is the displacement from equilibrium.  Us only depends on how much the spring is stretched or compressed, not on how this was accomplished.

Problem:

A 100 N force is used to stretch a spring with spring constant k = 2000 N/m.
(a)  What is the maximum stretch this force can produce?
(b)  How much elastic potential energy is stored in the spring at the maximum stretch?

Solution:
(a)  The spring obeys Hooke's law. 
At maximum stretch Fext = kx.
x = Fext/k = (100 N)/(2000 N/m) = 0.05 m = 5 cm.
(b)  The elastic potential energy stored in the spring is
Us= ½kx2 = ½(2000 N/m)(0.05 m)2 = 2.5 J.


When the gravitational force does work on an object near the surface of Earth, the potential energy of the object changes.  We may write

F∆x = -∆U or F = -∆U/∆x.

F = -∆U/∆x (as ∆x --> 0) holds for any conservative force in one dimension.  If we know the potential energy function for the object, we can calculate the conservative force acting on the object.

Using calculus:
For an isolated, conservative system E is constant, dE/dt = 0.  Assume an object is moving along the x-axis.
dK/dt = d(mv2/2)/dt = mv(dv/dt) = mvax = Fxv.
dU/dt = (dU/dx)(dx/dt) = (dU/dx)v.
dK/dt + dU/dt = 0 implies Fxv + (dU/dx)v = 0.
or
Fx = -dU/dx.
The conservative internal force acting between parts of the system equals the negative derivative of the potential energy associated with that system.