Elastic Potential Energy (Cambridge (CIE) A Level Physics): Revision Note

The graph shows the behaviour of a sample of a metal when it is stretched until it starts to undergo plastic deformation.

What is the total work done in stretching the sample from zero to an extension of 13.5 mm?

Simplify the calculation by treating the curve XY as a straight line.

Answer:

Step 1: Recall how to determine work done from the graph:

• Work done is the area underneath the force-extension graph

Step 2: Calculate the area under the graph up to point X:

• To point X, the area under the graph, A_X , is a triangle

$A_{\mathrm{X}}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

• Calculate A_X , remembering to convert length to metres

$A_{\mathrm{X}}=\frac{1}{2}\times \left(11.0\times {10}^{-3}\right)\times 450=2.475\mathrm{J}$

Step 3: Calculate the area between X and Y:

• Assuming the line XY is a straight line, the area under this region of the graph forms a trapezium

• Recall the equation for a trapezium of width h  and side lengths a  and b 

  $A_{\mathrm{XY}}=\left(\frac{a+b}{2}\right)h$

  • Here, h is the change in extension from X to Y, 2.5 mm

  • a  is the load at point X and b  is the load at point Y

$A_{\mathrm{XY}}=\left(\frac{450+600}{2}\right)\times \left(2.5\times {10}^{-3}\right)=1.313\mathrm{J}$

Step 4: Calculate total area:

• The total area, the total work done, is just the sum of these two areas

$\mathrm{work}\mathrm{done}=2.475+1.313=3.79\mathrm{J}$

• The answer is given to 3 significant figures, as the data has been given to this number of significant figures

Make sure to be familiar with the formula for the area of common 2D shapes such as a right angled triangle, trapezium, square and rectangles. If you do forget the equation for a trapezium's area, however, just split the shape up into rectangles and triangles.

A spring is extended with varying forces; the graph below shows the results.

What is the energy stored in the spring when the extension is 40 mm?

Answer:

Step 1: Recall how to determine energy sto

• Energy stored in the spring is equal to area under the graph, A

• This is a triangle, so can be calculated using

$A=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Step 2: Find the area under the graph

• At 40 mm, the load is approximately 8.3 N

• Converting length into metres gives an area of

$A=\frac{1}{2}\times \left(40\times {10}^{-3}\right)\times 8.3=0.166\mathrm{J}$

• This is equal to the energy stored in the spring, which should be given to 2 significant figures

$\mathrm{energy}\mathrm{stored}=0.17\mathrm{J}$

In your exam, a range of values will be allowed for the load at 40 mm. Any value from 8.1 N to 8.5 N will gain a mark. If you are struggling, draw horizontal lines on the graph to show the positions of 9.0 N and 8.5 N.

The formula for EPE = $\frac{1}{2}k{x}^2$ is only the area under the force-extension graph when it is a straight line i.e. when the material obeys Hooke’s law and is within its elastic limit.