A light elastic string has modulus of elasticity $$\(2 \mathrm{mg}\)$$ and natural length $$\(l\)$$. One end of the string is fixed to a point $$\(A\)$$ on a rough plane inclined to the horizontal at angle $$\(\alpha\)$$, where $$\(\sin \alpha=\frac{3}{5}\)$$. A particle, $$\(P\)$$, of mass $$\(m\)$$ is attached to the other end of the string. Initially $$\(P\)$$ is held at rest on the plane at the point $$\(B\)$$, where $$\(B\)$$ is above $$\(A\)$$ and $$\(A B=\frac{1}{2} l\)$$. The string lies along a line of greatest slope of the plane. The particle $$\(P\)$$ is released from rest and moves down the plane along the line of greatest slope. The coefficient of friction between $$\(P\)$$ and the plane is $$\(\mu\)$$, where $$\(\mu< \tan \alpha\)$$. Given that $$\(P\)$$ comes to instantaneous rest at the point $$\(C\)$$, where $$\(A C=l+e\)$$, (a) show that $$\[ \mu=\frac{9 l^{2}+6 l e-10 e^{2}}{4 l(3 l+2 e)} \]$$ (6) Given that $$\(e=l\)$$ (b) find the magnitude of the instantaneous change in the acceleration of $$\(P\)$$ at $$\(C\)$$. (5)

Answer:

Knowledge points:

Solution: