CAIE A-Level Physics 6.2 Elastic and Plastic Behaviour

Practise interpreting loading curves, elastic limits and permanent deformation, finding work from graph area and calculating stored elastic energy.

Exam points

distinguish elastic and plastic deformation and identify the elastic limit from a graph or from what happens when the load is removed
interpret the area under a force–extension graph as work done and calculate areas for simple or irregular graph regions
determine elastic potential energy from graph area or from EP = 1/2Fx = 1/2kx² within the limit of proportionality

Covered syllabus points

6.2.1—The terms elastic deformation, plastic deformation and elastic limit 6.2.2—The area under the force–extension graph represents the work done 6.2.3—The elastic potential energy of a material deformed within its limit of 6.2.4—EP = 1/2 Fx = 1/2 kx2 for a material deformed within its limit of

[Maximum number: 1]

One end of a wire is attached to a fixed point. A force F is applied to the wire to cause extension x. The variation with F of x is shown in Fig. 5.1.

Fig. 5.1

The wire has a cross-sectional area of $4.1 \times 10^{-7} \mathrm{~m}^{2}$ and is made of metal of Young modulus $1.7 \times 10^{11} \mathrm{~Pa}$ . Assume that the cross-sectional area of the wire remains constant as the wire extends.

A force of greater than 45 N is now applied to the wire.

Describe how it may be checked that the elastic limit of the wire has not been exceeded.

[Maximum number: 5]

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.

(a)

For the lines in Fig. 4.1:

[ 2 ]

(i)

explain why the area under the line represents the elastic potential energy of the wire.

[ 2 ]

(b)

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

[ 3 ]

(i)

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

E_{P}=

[ 3 ]

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

[Maximum number: 2]

A spring is fixed at one end and is compressed by applying a force to the other end. The variation of the force F acting on the spring with its compression x is shown in Fig. 3.1.

Fig. 3.1

A compression of 0.045 m is produced when a force $F_{1}$ acts on the spring. The spring has a spring constant of $800 \mathrm{Nm}^{-1}$ .

Use Fig. 3.1 to show that, for a compression of 0.045 m , the elastic potential energy of the spring is 0.81 J .

$(E=) \frac{1}{2} k x^{2}$ or $\frac{1}{2} F x$ or area under graph

C1

$\frac{1}{2} \times 800 \times(0.045)^{2}$ or $\frac{1}{2} \times 36 \times 0.045=0.81$ (J)

A1

All question bank results loaded

Covered syllabus points

6.2 Elastic and plastic behaviour question 1

6.2 Elastic and plastic behaviour question 2

Question (a)

Question (i)

Question (b)

Question (i)

6.2 Elastic and plastic behaviour question 3