Question
A bungee jumper of mass \(64\,\mathrm{kg}\) secures one end of an elastic rope to a bridge. The other end is attached to the jumper, who falls from the bridge and descends into the valley below, as shown in Fig. 3.1.
Fig. 3.2 shows the variation of the tension \(T\) in the rope with the vertical distance \(h\) of the jumper below the level of the bridge.
(a) The rope obeys Hooke’s law.
State Hooke’s law. (1 mark)
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(b) (i) Determine the unstretched length of the rope. (1 mark)
length = __________________________ \( \mathrm{m} \)
(ii) Determine the spring constant \(k\) of the rope. (2 marks)
\(k=\) __________________________ \( \mathrm{N\,m^{-1}} \)
(c) For the position of the bungee jumper at a distance of \(120\,\mathrm{m}\) below the bridge:
(i) Show that the loss of gravitational potential energy since leaving the bridge is \(75\,\mathrm{kJ}\). (2 marks)
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(ii) Show that the elastic potential energy in the rope is \(75\,\mathrm{kJ}\). (2 marks)
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____________________________________________
(d) Explain what can be deduced from the information in (c) about the speed of the bungee jumper at a distance of \(120\,\mathrm{m}\) below the bridge. (2 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 5.1: Work, energy and power — parts (c)(i), (c)(ii) and (d)
▶️ Answer/Explanation
(a) Hooke’s law [1 mark]
Hooke’s law states that the force is proportional to the extension, provided the limit of proportionality is not exceeded.
Answer: \( \boxed{F\propto x} \)
(b)(i) Unstretched length [1 mark]
The rope begins to experience tension when the jumper has fallen to the unstretched length of the rope.
From Fig. 3.2, the tension becomes non-zero at approximately \(h=37\,\mathrm{m}\).
Answer: \( \boxed{37\,\mathrm{m}} \)
(b)(ii) Spring constant [2 marks]
Using Hooke’s law,
\(T=kx\)
At \(h=120\,\mathrm{m}\), the extension is
\(x=120-37=83\,\mathrm{m}\)
The tension is approximately \(1800\,\mathrm{N}\).
\(k=\dfrac{1800}{120-37}\)
\(k=21.7\,\mathrm{N\,m^{-1}}\)
Answer: \( \boxed{22\,\mathrm{N\,m^{-1}}} \)
(c)(i) Loss of gravitational potential energy [2 marks]
The loss of gravitational potential energy is
\(\Delta E_{\mathrm{P}}=mg\Delta h\)
\(\Delta E_{\mathrm{P}}=64\times9.81\times120\)
\(\Delta E_{\mathrm{P}}=75000\,\mathrm{J}\)
\(75000\,\mathrm{J}=75\,\mathrm{kJ}\)
Answer: \( \boxed{75\,\mathrm{kJ}} \)
(c)(ii) Elastic potential energy [2 marks]
The elastic potential energy stored in the rope is
\(E=\dfrac{1}{2}Fx\)
Using \(F=1800\,\mathrm{N}\) and \(x=120-37=83\,\mathrm{m}\),
\(E=\dfrac{1}{2}\times1800\times(120-37)\)
\(E=75000\,\mathrm{J}\)
\(E=75\,\mathrm{kJ}\)
Answer: \( \boxed{75\,\mathrm{kJ}} \)
(d) Speed of the bungee jumper [2 marks]
The loss of gravitational potential energy is \(75\,\mathrm{kJ}\), while the elastic potential energy stored in the rope is also \(75\,\mathrm{kJ}\).
Therefore, all of the gravitational potential energy lost has been converted into elastic potential energy.
There is no remaining energy available as kinetic energy, so the kinetic energy of the jumper is zero.
Answer: \( \boxed{\text{the speed of the jumper is zero}} \)