I have a question that reads:

A light elastic string $AB$, of natural length $1.2$ m is fixed at point $A$ on a rough plane inclined at $30^\circ$ to the horizontal. The string had modulus of elasticity $115$ N. A particle of mass $2$ kg is attached to end $B$ and the particle is released from rest to descend the plane from A to C. The particle descends $1.45$ m from $A$.

Show that the coefficient of friction between the particle and inclined plane is $0.456$.

A worked out solution uses the conservation of energy: $$ 2g \times \sin 30^\circ \times 1.45 - \mu \times 2g \times \cos 30^\circ \times 1.45 = \frac{115 \times 0.25^2}{2 \times 1.2} $$

I feel like the worked solution assumes the kinetic energy $1.45$ m down is zero.

It looks like they've done: $$ \text{Work done by gravity} \\ - \text{Work done against friction} \\ - \text{Work done against tension} \\ = \text{increase in KE (0)} $$ But nowhere in the question does it say that the particle is stationary at C (1.45m down). Perhaps its implied but I just want to make sure I'm not missing something vitally important. Equally I don't know the speed at C so I'm not sure how to work out the question otherwise.


You are correct in the fact that the particle is at rest at point $C$, the velocity is then $0$. Otherwise you don't have enough information to solve the problem. I think you have an error in your formula, in the term on the right hand side. There should not be $1.2$ in the denominator. The work done by the elastic force is $k(AC-AB)^2/2$.

  • $\begingroup$ But shouldn't the denominator be twice the natural length of the string? $\endgroup$
    – Gab N.
    Dec 27 '18 at 15:56
  • $\begingroup$ The modulus of elasticity should be $N/m$, so you might be right. But that's definitely not how the modulus of elasticity should be defined. By definition it should be $k=F/\Delta l$. $\Delta l$ is a change in length. The formula should not contain the original non-stretched length of the string. $\endgroup$
    – Andrei
    Dec 27 '18 at 16:10

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