FM1-3 Elastic strings and springs and elastic energy — coverage pack

2 specification leaves · notes, questions, answers and worked methods

FM1-3.1 · Elastic strings and springs. Hooke's law.

For a spring obeying Hooke's law, the tension has magnitude $T=kx$ , where $k$ is the spring constant and $x$ is the extension.
For an elastic string of natural length $l$ and modulus $\lambda$ , Hooke's law is $T=\lambda x/l$ while the string is taut.
Extension is stretched length minus natural length. A string becomes slack rather than exerting a compressive force when this value is non-positive.
Keep spring constant and modulus distinct: $k$ has units $\text{N m}^{-1}$ , whereas $\lambda$ has units $\text{N}$ .

1. A light spring has spring constant $80\,\text{N m}^{-1}$ and is extended by $0.06\,\text{m}$ . Find its tension.[2 marks]
Answer
- $4.8\,\text{N}$
Method: Hooke's law for a spring gives $T=kx=80(0.06)=4.8\,\text{N}$ .

1. An elastic string has natural length $1.5\,\text{m}$ and modulus $120\,\text{N}$ . It is stretched to length $1.8\,\text{m}$ . Calculate its tension.[3 marks]
Answer
- $24\,\text{N}$
Method: The extension is $x=1.8-1.5=0.3\,\text{m}$ . Therefore $T=\lambda x/l=120(0.3)/1.5=24\,\text{N}$ .

1. A $1.5\,\text{kg}$ particle hangs in equilibrium from a vertical elastic string of natural length $0.8\,\text{m}$ and modulus $49\,\text{N}$ . Taking $g=9.8\,\text{m s}^{-2}$ , determine the extension and the equilibrium length of the string.[4 marks]
Answer
- Extension $0.24\,\text{m}$
- Equilibrium length $1.04\,\text{m}$
Method: Equilibrium gives $T=mg=1.5(9.8)=14.7\,\text{N}$ . Hooke's law gives $14.7=49x/0.8$ , so $x=0.24\,\text{m}$ . Adding the natural length gives $0.8+0.24=1.04\,\text{m}$ .

A spring of constant $k$ stores elastic energy $\tfrac12kx^2$ when extended or compressed by $x$ .
A taut elastic string of natural length $l$ and modulus $\lambda$ stores energy $\lambda x^2/(2l)$ at extension $x$ .
Elastic energy is the area under the tension-extension graph, so it is quadratic in extension rather than equal to final tension times extension.
In an energy equation, include elastic energy only while the string is taut, and measure extension from the natural length at each position.

1. A spring of constant $200\,\text{N m}^{-1}$ is compressed by $0.08\,\text{m}$ . Calculate the elastic energy stored.[2 marks]
Answer
- $0.64\,\text{J}$
Method: $E=\tfrac12kx^2=\tfrac12(200)(0.08)^2=0.64\,\text{J}$ .

1. An elastic string has natural length $1.2\,\text{m}$ and modulus $60\,\text{N}$ . Find the energy stored when its length is $1.5\,\text{m}$ .[3 marks]
Answer
- $2.25\,\text{J}$
Method: The extension is $x=1.5-1.2=0.3\,\text{m}$ . Thus $E=\lambda x^2/(2l)=60(0.3)^2/[2(1.2)]=2.25\,\text{J}$ .

1. A $2\,\text{kg}$ particle is attached to the lower end of a vertical elastic string with natural length $1\,\text{m}$ and modulus $49\,\text{N}$ . It is released from rest when the string is at natural length. Taking $g=9.8\,\text{m s}^{-2}$ , find its speed after descending $0.6\,\text{m}$ .[5 marks]
Answer
- $\sqrt{2.94}\,\text{m s}^{-1}\approx1.71\,\text{m s}^{-1}$
Method: The loss of gravitational potential energy is $2(9.8)(0.6)=11.76\,\text{J}$ . At extension $0.6\,\text{m}$ , elastic energy is $49(0.6)^2/[2(1)]=8.82\,\text{J}$ . Conservation of mechanical energy gives $11.76=\tfrac12(2)v^2+8.82$ , so $v^2=2.94$ and $v=\sqrt{2.94}\,\text{m s}^{-1}$ .