Physics 9702 · Paper 2

Elastic Potential Energy

Stretching a spring does work, and that work is stored as elastic potential energy $E_p$ (strain energy). The small $p$ marks it as potential energy, not the Young modulus $E$ .
There is one idea to hold on to: elastic PE is the area under the force-extension graph. From that, two formulae follow, one step at a time.

Energy is the area under the graph

E_p=\tfrac12 Fx

Symbols

$E_p$ = elastic potential energy, or strain energy (J)
$F$ = final force stretching the spring (N)
$x$ = extension (m)

The force increases from 0 to $F$ as the extension increases from 0 to $x$ , so the average force is $\tfrac12F$ .
Work done is average force times extension, $E_p=\tfrac12Fx$ : the area of the triangle under a straight line through the origin.
For a curved graph the area is still the energy, but you must find it by counting squares, not by $\tfrac12Fx$ .

The energy is the area under the force-extension graph. For a straight line that area is the triangle $\tfrac12Fx$ ; for a curve, measure the area directly.

A drawn bow with the string pulled taut and an arrow ready to fly, storing energy in the bent bow and stretched string. — Stretching stores energy. In this drawn bow, the work done pulling the string is stored as elastic potential energy; it is released when the arrow is shot.

Writing it with k instead of F

Often you know the spring constant $k$ and the extension $x$ , but not the force. Since $F=kx$ , substitute it into $\tfrac12Fx$ :

E_p=\tfrac12 kx^2

Symbols

$k$ = spring constant (N m⁻¹)
$x$ = extension (m)

Use whichever form fits the numbers you have: $\tfrac12Fx$ when you know the force, $\tfrac12kx^2$ when you know $k$ .

Worked example

Elastic PE stored in a spring

A spring of constant 2800 N m^-1 is stretched by 0.21 m.

Use $E_p=\tfrac12kx^2$ .
$E_p=\tfrac12\times2800\times0.21^2=62\text{ J}$ .

Answer

E_p=62\text{ J}

(9702/21/M/J/25 Q2(c))

Why the area is the energy, and two exam variants

Work done is force times distance. As the force changes with extension, split the stretch into thin strips, each of work $F\,\Delta x$ ; adding them is exactly the area under the graph. (9702/23/O/N/23 Q4)
Because $E_p=\tfrac12kx^2$ , the stored energy scales with $x^2$ . Comparing two stretches of the same spring gives a ratio, for example $E_1/E_2=2.0$ . (9702/22/O/N/25 Q3(c)(ii))
To find the work done stretching from $x_1$ to $x_2$ , take the difference of the stored energies, $\tfrac12k(x_2^2-x_1^2)$ (the trapezium area), not $\tfrac12F(x_2-x_1)$ . (9702/22/M/J/25 Q5(b)(ii))

PreviousMeasuring the Young Modulus NextThe Stress-Strain Graph

Elastic Potential Energy

Stretching a spring does work, and that work is stored as elastic potential energy $E_p$ (strain energy). The small $p$ marks it as potential energy, not the Young modulus $E$ .
There is one idea to hold on to: elastic PE is the area under the force-extension graph. From that, two formulae follow, one step at a time.

Energy is the area under the graph

E_p=\tfrac12 Fx

Symbols

$E_p$ = elastic potential energy, or strain energy (J)
$F$ = final force stretching the spring (N)
$x$ = extension (m)

The force increases from 0 to $F$ as the extension increases from 0 to $x$ , so the average force is $\tfrac12F$ .
Work done is average force times extension, $E_p=\tfrac12Fx$ : the area of the triangle under a straight line through the origin.
For a curved graph the area is still the energy, but you must find it by counting squares, not by $\tfrac12Fx$ .

The energy is the area under the force-extension graph. For a straight line that area is the triangle $\tfrac12Fx$ ; for a curve, measure the area directly.

Writing it with k instead of F

Often you know the spring constant $k$ and the extension $x$ , but not the force. Since $F=kx$ , substitute it into $\tfrac12Fx$ :

E_p=\tfrac12 kx^2

Symbols

$k$ = spring constant (N m⁻¹)
$x$ = extension (m)

Use whichever form fits the numbers you have: $\tfrac12Fx$ when you know the force, $\tfrac12kx^2$ when you know $k$ .

Worked example

Elastic PE stored in a spring

A spring of constant 2800 N m^-1 is stretched by 0.21 m.

Use $E_p=\tfrac12kx^2$ .
$E_p=\tfrac12\times2800\times0.21^2=62\text{ J}$ .

Answer

E_p=62\text{ J}

(9702/21/M/J/25 Q2(c))

Why the area is the energy, and two exam variants

Work done is force times distance. As the force changes with extension, split the stretch into thin strips, each of work $F\,\Delta x$ ; adding them is exactly the area under the graph. (9702/23/O/N/23 Q4)
Because $E_p=\tfrac12kx^2$ , the stored energy scales with $x^2$ . Comparing two stretches of the same spring gives a ratio, for example $E_1/E_2=2.0$ . (9702/22/O/N/25 Q3(c)(ii))
To find the work done stretching from $x_1$ to $x_2$ , take the difference of the stored energies, $\tfrac12k(x_2^2-x_1^2)$ (the trapezium area), not $\tfrac12F(x_2-x_1)$ . (9702/22/M/J/25 Q5(b)(ii))