Elastic Potential Energy | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Elastic Potential Energy

Stretching or compressing an object requires work because a force acts through a distance.

While the deformation is elastic, this work is stored as recoverable elastic potential energy.

A hand-drawn bow is bent while its string is pulled back with an arrow ready to be released.

work done while pullingelastic energy stored

The pulling force does work while the string moves. While the bow remains elastically deformed, that transferred energy can be recovered when it returns to its original shape.

The area under the graph is the transferred energy

The force is not constant while a Hooke's-law spring is stretched. It rises from zero to its final value. Therefore, multiplying the final force by the extension would count too much work.

Swipe left or right to see the whole diagram →

For a Hooke's-law region, the force rises from zero to F. The triangular area is ½ × F × x, so it equals the elastic potential energy stored.

E_{\mathrm p}=\frac12 Fx

Symbols

$E_{\mathrm p}$ = elastic potential energy (J)
$F$ = final force (N)
$x$ = extension (m)

The factor one-half comes from the triangular area: the average force during the stretch is half the final force. This formula applies to the straight Hooke's-law region.

E_{\mathrm p}=\frac12 kx^2

Symbols

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

This second form follows by substituting $F=kx$ . It also applies only while Hooke's law is obeyed.

Change extension

k = 200 N m⁻¹ · x = 0.040 m · F = 8.0 N · elastic PE = 0.160 J

Your turn9702/21/M/J/25 Q2(c)

A spring has spring constant 2800 N m^-1 and extension 0.21 m. Calculate the elastic potential energy stored. (9702/21/M/J/25 Q2(c))

Show worked answer

Use $E_{\mathrm p}=\tfrac12kx^2$ .
$E_{\mathrm p}=\tfrac12(2800)(0.21)^2=61.7\text{ J}$ .
Give two significant figures to match the data.

Answer

E_{\mathrm p}=62\text{ J}

Common mistake

Do not use

Fx

for a spring that starts unstretched. The force rises from zero, so the graph area is a triangle and contains the factor

\tfrac12

Elastic Potential Energy

Stretching or compressing an object requires work because a force acts through a distance.

While the deformation is elastic, this work is stored as recoverable elastic potential energy.

work done while pullingelastic energy stored

The pulling force does work while the string moves. While the bow remains elastically deformed, that transferred energy can be recovered when it returns to its original shape.

The area under the graph is the transferred energy

The force is not constant while a Hooke's-law spring is stretched. It rises from zero to its final value. Therefore, multiplying the final force by the extension would count too much work.

Swipe left or right to see the whole diagram →

For a Hooke's-law region, the force rises from zero to F. The triangular area is ½ × F × x, so it equals the elastic potential energy stored.

E_{\mathrm p}=\frac12 Fx

Symbols

$E_{\mathrm p}$ = elastic potential energy (J)
$F$ = final force (N)
$x$ = extension (m)

The factor one-half comes from the triangular area: the average force during the stretch is half the final force. This formula applies to the straight Hooke's-law region.

E_{\mathrm p}=\frac12 kx^2

Symbols

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

This second form follows by substituting $F=kx$ . It also applies only while Hooke's law is obeyed.

Change extension

k = 200 N m⁻¹ · x = 0.040 m · F = 8.0 N · elastic PE = 0.160 J

Your turn9702/21/M/J/25 Q2(c)

A spring has spring constant 2800 N m^-1 and extension 0.21 m. Calculate the elastic potential energy stored. (9702/21/M/J/25 Q2(c))

Show worked answer

Use $E_{\mathrm p}=\tfrac12kx^2$ .
$E_{\mathrm p}=\tfrac12(2800)(0.21)^2=61.7\text{ J}$ .
Give two significant figures to match the data.

Answer

E_{\mathrm p}=62\text{ J}

Common mistake

Do not use

Fx

for a spring that starts unstretched. The force rises from zero, so the graph area is a triangle and contains the factor

\tfrac12