# Assuming a particle is at rest/stationary

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 $$k(AC-AB)^2/2$$.
• 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. Dec 27 '18 at 16:10