CAIE A-Level Physics 6.2.4 Elastic Energy Formulae

Practise using EP = ½Fx or EP = ½kx² for elastic energy, rearranging for force or extension and checking the limit of proportionality.

Exam points

calculate elastic potential energy using EP = 1/2Fx or EP = 1/2kx² within the limit of proportionality
rearrange the elastic-energy formula to determine force, extension or spring constant
compare energy stored for different forces, extensions or spring constants while checking that the stated deformation remains within the valid range

[Maximum number: 3]

Fig. 4.1 shows the variation with extension x of the tensile force F for two wires, G and H, made from the same material.

Fig. 4.1

The elastic limit has not been exceeded for G or H.

Wires G and H are joined together end-to-end to form a composite wire of negligible weight. The composite wire hangs vertically from a fixed support.

A block of weight of 2.0 N is attached to the end of the wire, as shown in Fig. 4.2.

Fig. 4.2

Calculate the total elastic potential energy $E_{\mathrm{P}}$ of the composite wire due to the weight of the block.

E_{P}=

E=1 / 2 F x

or

$E=1 / 2 k x^{2}$ and F=k x

or

E= area under graph

C1

$\begin{aligned} \text { for } \mathrm{G}: E=1 / 2 \times 2.0 \times\left(0.39 \times 10^{-3}\right) \text { or } 1 / 2 \times 5.1 \times 10^{3} \times\left(0.39 \times 10^{-3}\right)^{2} \text { for } \mathrm{H}: E=1 / 2 \times 2.0 \times\left(0.29 \times 10^{-3}\right) \text { or } 1 / 2 \times 6.9 \times 10^{3} \times\left(0.29 \times 10^{-3}\right)^{2} \begin{aligned} E_{\mathrm{P}}= 3.9 \times 10^{-4}+2.9 \times 10^{-4} =6.8 \times 10^{-4} \mathrm{~J} \end{aligned} \end{aligned}$

A1

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6.2.4—EP = 1/2 Fx = 1/2 kx2 for a material deformed within its limit of question 1