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S3.6 Potential energy: examples

[A] Linear (spring) force: potential energy

Key Point 3.12

The potential energy stored in a spring with displacement x is

\[ U = \frac{1}{2} k x^2 \]

Note: convention dictates we choose the reference point to be the displacement from the equilibrium length.
Graphic - No title - hookeslaw

Analysis

Analysis

Since in one dimension

\[ F = -\frac{dU}{dx} \]
we have that the restoring force for a stretched or compressed spring is

F(x)=-kx
which is known as Hooke’s law.

 

Worked example

Worked example

The scale of a certain spring balance reads from zero to 1200N and is 0.1m long.

What is the potential energy of the spring when :

  • [(a)] it is stretched 0.1m? 0.05m?
  • [(b)] a 60kg mass hangs from the spring.

Solution

A force of -1200N fully stretches the spring to 0.1m. Thus, using Hooke’s law, the spring constant is k=-F/x=1200/0.1=1.2×104N/m.

  • [(a)] The potential energy for any extension is given by U(x)=kx2/2. Thus one finds U(0.1)=60J, U(0.05)=15J.
  • [(b)] For a 60kg mass the extension will be x=mg/k=0.05m and hence the potential energy is again U(0.05)=15J.
 

[B] Gravitational potential energy of a body near the earth’s surface

Note: the reference point we choose is arbitrary

Analysis

Analysis

We can check this by calculating

\[ F = -\frac{dU}{dx} \;, \]
with U(x)=mgx. Hence

F(x)=-mg
which is, of course, just the weight of the body.

 

[C] Gravitational potential energy of two point masses

Analysis

Analysis

calculating the associated force as before, we find

\[ F = \frac{Gm_1m_2}{r^2} \;, \]

the attractive gravitational force between two bodies. (S2.10)

Note: By definition, two bodies infinitely far apart have a potential energy of zero: this is the reference point, infinite separation.

 
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TTwo Planets and a Comet

Learning Resources

Textbook: HRW Chapter 8.4
Course Questions:
Self-Test Questions: