# CIE AS & A Level Physics : 6.1 Stress and strain – Exam style question – Paper 2

### Question

(a) A uniform metal bar, initially unstretched, has sides of length w, x and y, as shown in Fig. 3.1.

The bar is now stretched by a tensile force F applied to the shaded ends. The changes in
the lengths x and y are negligible. The bar now has sides of length x, y and z, as shown in
Fig. 3.2.

Determine expressions, in terms of some or all of F, w, x, y and z, for:
(i) the stress σ applied to the bar by the tensile force
σ = ………………………………………………… [1]
(ii) the strain ε in the bar due to the tensile force
ε = ………………………………………………… [1]
(iii) the Young modulus E of the metal from which the bar is made.
E = ………………………………………………… [2]

(b)  A copper wire is stretched by a tensile force that gradually increases from 0 to 280N. The
variation with extension of the tensile force is shown in Fig. 3.3.

(i) State the maximum extension of the wire for which it obeys Hooke’s law.
extension = ………………………………………….. mm [1]
(ii) Use Fig. 3.3 to determine the strain energy in the wire when the tensile force is 120N.
strain energy = ……………………………………………… J [3]
(iii) Explain why the work done in stretching the wire to an extension of 12mm is not equal to
the energy recovered when the tensile force is removed.                                                    [2]
[Total: 10]

(a) (i) σ = F / xy
(a) (ii) ε = (z –w) / w
(a) (iii) E =σ / ε
= Fw / xy(z–w)
(b) (i) extension = 2.2 mm (allow 2.0 –2.4 mm)
(b) (ii) strain energy = area under graph/line or ½Fx or ½kx2
= ½ × 120 × 1.4 × 10–3 or ½ × 8.6 × 104 × (1.4 × 10–3)
= 0.084 J
(b) (iii) (some of the) deformation of the wire is plastic/permanent/not elastic
or
wire goes past the elastic limit/enters plastic region
energy (that cannot be recovered) is dissipated as thermal energy/becomes internal energy

### Question

A spring is extended by a force. The variation with extension x of the force F is shown in Fig. 3.1

(a) State the name of the law that relates the force and extension of the spring shown in Fig. 3.1.     [1]
(b) Determine:
(i) the spring constant, in Nm−1, of the spring
spring constant = ……………………………………….. Nm−1 [2]
(ii) the strain energy (elastic potential energy) in the spring when the extension is 4.0cm.
strain energy = ……………………………………………… J [2]

(c) One end of the spring is attached to a fixed point. A cylinder that is submerged in a liquid is
now suspended from the other end of the spring, as shown in Fig. 3.2.

The cylinder has length 5.8cm, cross-sectional area 1.2 × 10−3m2 and weight 6.20N. The
cylinder is in equilibrium when the extension of the spring is 4.0cm.
(i) Show that the upthrust acting on the cylinder is 0.60N.                                                            [1]
(ii) Calculate the difference in pressure between the bottom face and the top face of the
cylinder.
difference in pressure = ……………………………………………. Pa [2]

(iii) Calculate the density of the liquid.
density = ………………………………………. kgm−3 [2]
(d) The liquid in (c) is replaced by another liquid of greater density.
State the effect, if any, of this change on:
(i) the upthrust acting on the cylinder                                                                                               [1]
(ii) the extension of the spring.                                                                                                           [1]
[Total: 12]

Ans

(a) Hooke’s (law)

(b) (i) k = F / x or k = gradient
e.g. k = 7.0 / 5.0 × 10–2
= 140 N m–1

(b) (ii) E = ½ F x or E = ½ k x2 or E = area under graph
= ½ × 5.6 × 4.0 × 10–2 or ½ × 140 × (4.0 × 10–2)2
= 0.11 J

(c) (i) (upthrust =) 6.20 – 5.60 = 0.60 (N)

(c) (ii)Δp =Δ F / A
= 0.60 / 1.2 × 10–3
= 500 Pa

(c) (iii) (Δ)p = ρg(Δ)h
ρ = 500 / (9.81 × 5.8 × 10–2)
= 880 kg m–3

(d) (i) (upthrust) increases

(d) (ii) (extension) decreases

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