2.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-04_401_1031_287_516}
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\caption{Figure 1}
\end{figure}
A light elastic spring has natural length \(l\) and modulus of elasticity \(\lambda\) One end of the spring is attached to a point \(A\) on a smooth plane.
The plane is inclined at angle \(\theta\) to the horizontal, where \(\tan \theta = \frac { 5 } { 12 }\) A particle \(P\) of mass \(m\) is attached to the other end of the spring.
Initially \(P\) is held at the point \(B\) on the plane, where \(A B\) is a line of greatest slope of the plane.
The point \(B\) is lower than \(A\) and \(A B = 2 l\), as shown in Figure 1 .
The particle is released from rest at \(B\) and first comes to instantaneous rest at the point \(C\) on \(A B\), where \(A C = 0.7 l\)
- Use the principle of conservation of mechanical energy to show that
$$\lambda = \frac { 100 } { 91 } m g$$
- Find the acceleration of \(P\) when it is released from rest at \(B\).