15 In this question use \(g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
A particle \(P\) of mass \(m\) is attached to two light elastic strings, \(A P\) and \(B P\).
The other ends of the strings, \(A\) and \(B\), are attached to fixed points which are 4 metres apart on a rough horizontal surface at the bottom of a container.
The coefficient of friction between \(P\) and the surface is 0.68
- When the extension of string \(A P\) is \(e _ { A }\) metres, the tension in \(A P\) is \(24 m e _ { A }\)
- When the extension of string \(B P\) is \(e _ { B }\) metres, the tension in \(B P\) is \(10 m e _ { B }\)
- The natural length of string \(A P\) is 1 metre
- The natural length of string \(B P\) is 1.3 metres
\includegraphics[max width=\textwidth, alt={}, center]{8f7a5fc0-6936-4aed-a173-e221bf86e4fd-24_92_1082_1030_479}
15
- Show that when \(A P = 1.5\) metres, the tension in \(A P\) is equal to the tension in \(B P\).
15 - \(\quad P\) is held at the point between \(A\) and \(B\) where \(A P = 1.9\) metres, and then released from rest.
At time \(t\) seconds after \(P\) is released, \(A P = ( 1.5 + x )\) metres.
\includegraphics[max width=\textwidth, alt={}, center]{8f7a5fc0-6936-4aed-a173-e221bf86e4fd-25_140_1068_493_484}
Show that when \(P\) is moving towards \(A\),
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 34 x = 6.664$$
15 - The container is then filled with oil, and \(P\) is again released from rest at the point between \(A\) and \(B\) where \(A P = 1.9\) metres.
At time \(t\) seconds after \(P\) is released, the oil causes a resistive force of magnitude \(10 m v\) newtons to act on the particle, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the particle.
Find \(x\) in terms of \(t\) when \(P\) is moving towards \(A\).
\includegraphics[max width=\textwidth, alt={}, center]{8f7a5fc0-6936-4aed-a173-e221bf86e4fd-27_2492_1721_217_150}
\includegraphics[max width=\textwidth, alt={}]{8f7a5fc0-6936-4aed-a173-e221bf86e4fd-32_2486_1719_221_150}