Define the stiffness of a spring.
Sign in and upgrade to Premium to write and mark your own answer.
the force per unit extension; produced when the spring obeys Hooke’s law / within the limit of proportionality
Hookes Law And Deformation
A wire obeys Hooke’s law and has a stiffness of 4.0 × 10³ N m⁻¹. Calculate the force needed to stretch it by 2.5 mm.
Sign in and upgrade to Premium to write and mark your own answer.
Converting the extension to metres $$\Delta x=2.5\times {10}^{-3}\text{ m}$$ Calculating the force $$\Delta F=k\Delta x=4.0\times {10}^{3}\times 2.5\times {10}^{-3}$$ $$\Delta F=10\text{ N}$$
A spring extends by 4.0 cm when a force of 6.0 N is applied. The force is then doubled. Predict the new extension and state the assumption made.
Sign in and upgrade to Premium to write and mark your own answer.
extension doubles to 8.0 cm; because force is directly proportional to extension; the spring still obeys Hooke’s law / the limit of proportionality is not exceeded
State why strain has no units.
Sign in and upgrade to Premium to write and mark your own answer.
it is a ratio of two lengths / change in length divided by original length
A steel cable of cross-sectional area 1.5 × 10⁻⁴ m² supports a load of 4500 N. Calculate the tensile stress in the cable.
Sign in and upgrade to Premium to write and mark your own answer.
$$\sigma =\frac{F}{A}=\frac{4500}{1.5\times {10}^{-4}}$$ $$\sigma =3.0\times {10}^{7}\text{ Pa}$$
Show that a copper wire of length 1.80 m and cross-sectional area 2.0 × 10⁻⁷ m² extends by approximately 0.81 mm when a force of 10.0 N is applied. The Young modulus of copper is 1.1 × 10¹¹ Pa.
Sign in and upgrade to Premium to write and mark your own answer.
$$\Delta L=\frac{F{L}_{0}}{AE}$$ $$\Delta L=\frac{10.0\times 1.80}{2.0\times {10}^{-7}\times 1.1\times {10}^{11}}$$ $$\Delta L=\frac{18}{22000}=8.18\times {10}^{-4}\text{ m}$$ This is approximately 0.81 mm, as required.
Two wires P and Q are made of the same material. Wire Q has twice the diameter and twice the length of wire P. The same force is applied to each wire. Compare the extensions of the two wires.
Sign in and upgrade to Premium to write and mark your own answer.
Q has 4× the cross-sectional area of P, so Q has ¼ the stress of P; Q has 2× the original length, so for the same strain Q would extend 2× as much; combining these, Q extends ½ as much as P (extension of Q = ¼ × 2 = ½ × extension of P)
Define the limit of proportionality.
Sign in and upgrade to Premium to write and mark your own answer.
the point beyond which the extension of the object is no longer directly proportional to the applied force / Hooke’s law no longer applies
Distinguish between elastic deformation and plastic deformation.
Sign in and upgrade to Premium to write and mark your own answer.
elastic deformation: the object returns to its original length when the load is removed; plastic deformation: the object retains a permanent change in length / does not return to its original length when the load is removed
Sketch a force-extension graph for a metal wire stretched until it breaks. Label the limit of proportionality, the elastic limit, the yield point, and the breaking point.
Sign in and upgrade to Premium to write and mark your own answer.
straight line from origin (1 mark); limit of proportionality at end of straight line (1 mark); elastic limit shortly after, with curve continuing (1 mark); yield point where curve flattens / large extension for small force, with breaking point at end of curve labelled (1 mark)
Define breaking stress.
Sign in and upgrade to Premium to write and mark your own answer.
the maximum stress a material can withstand before it fractures / the stress at which the material breaks
Show that the breaking force of a copper wire of cross-sectional area 1.2 × 10⁻⁶ m² is approximately 250 N. The breaking stress of copper is 2.1 × 10⁸ Pa.
Sign in and upgrade to Premium to write and mark your own answer.
$$F=\sigma A=2.1\times {10}^{8}\times 1.2\times {10}^{-6}$$ $$F=252\text{ N}\approx 250\text{ N}$$ as required.
A nylon thread of cross-sectional area 0.80 mm² has a breaking stress of 7.5 × 10⁷ Pa. Deduce whether the thread can safely support a 50 N load.
Sign in and upgrade to Premium to write and mark your own answer.
Converting the area $$A=0.80\times {10}^{-6}=8.0\times {10}^{-7}{\text{ m}}^{2}$$ Calculating the stress $$\sigma =\frac{F}{A}=\frac{50}{8.0\times {10}^{-7}}$$ $$\sigma =6.25\times {10}^{7}\text{ Pa}$$ This is less than the breaking stress of 7.5 × 10⁷ Pa, so the thread can safely support the load.
Discuss how the stress-strain graphs of a brittle material and a ductile material differ, and explain how each graph can be used to determine the Young modulus and the breaking stress of the material. In your answer you should describe the shape of each graph, explain how the Young modulus is found from each graph, and explain how the breaking stress is identified on each graph.
Sign in and upgrade to Premium to write and mark your own answer.
brittle materials show a nearly straight line from origin that ends abruptly at the fracture point with little or no plastic deformation; ductile materials show a straight line up to the limit of proportionality, then a curve through yield point and an extended plastic region before fracture; in both cases, the Young modulus equals the gradient of the linear region; for a brittle material the breaking stress is the stress at the fracture point at the end of the straight section; for a ductile material the breaking stress is the highest point (peak) of the curve, which may occur before the final fracture point; the area under the stress-strain curve up to fracture indicates the toughness, which is greater for ductile materials because of their large plastic region