Hooke's law states that the tension in a stretched string of natural length \(l\) and modulus of elasticity \(\lambda\) is \(\frac { \lambda x } { l }\) when its extension is \(x\).
Using this formula, prove that the work done in stretching a string from an unstretched position to a position in which its extension is \(e\) is \(\frac { \lambda e ^ { 2 } } { 2 l }\).
(3 marks)
A particle, of mass 5 kg , is attached to one end of a light elastic string of natural length 0.6 metres and modulus of elasticity 150 N . The other end of the string is fixed to a point \(O\).
Find the extension of the elastic string when the particle hangs in equilibrium directly below \(O\).
The particle is pulled down and held at the point \(P\), which is 0.9 metres vertically below \(O\).
Show that the elastic potential energy of the string when the particle is in this position is 11.25 J .
The particle is released from rest at the point \(P\). In the subsequent motion, the particle has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when it is \(x\) metres above \(\boldsymbol { P }\).
Show that, while the string is taut,
$$v ^ { 2 } = 10.4 x - 50 x ^ { 2 }$$
Find the value of \(x\) when the particle comes to rest for the first time after being released, given that the string is still taut.