Physics - W.E.P. Question with Solution | TestHub

Underestimate the work done against friction, and compare it with the initial gravitational potential energy of the eraser.

First of all, investigate whether the rubber eraser will start moving at all. It will do so provided that $mg\text{sin}\alpha >\mu mg\text{cos}\alpha$, i.e. $\mu <\text{tan}\alpha =\text{tan}45^o=1$. This is clearly the case, since $\mu =0.6$. So the eraser will start moving. (If $\mu =\sqrt{2}>1=\text{tan}45^o$ the eraser would not even begin to slide, so (D) is correct.)

The determination of how the normal force acting on the eraser varies with position is difficult, but it is certain that the frictional force is always larger than its initial value of $\mu mg\text{cos}\alpha =\mu mg\text{cos}45^o$. This is because, after the initial release, the angle with the horizontal made by the slope on which the eraser moves decreases, and, in addition, the track has to provide a centripetal force for the moving eraser. The path to reach the lowest point of the track would be one-eighth of a circle, with a length of $R\pi /4$. The work done against friction can now be underestimated by taking the normal force as if it always had its initial value:

$$W_f>W_{und}=\frac{\mu mg}{\sqrt{2}}\cdot \frac{\pi R}{4}=\frac{0.6mg\pi R}{4\sqrt{2}}\approx 0.333mgR$$

so (B) is correct and (A) is false. The gravitational potential energy difference between the initial position and the bottom of the track is $\Delta E_p=mgR(1-\text{cos}\alpha )=mgR(1-\text{cos}45^o)\approx 0.293mgR$. It can be seen that $\left|W_f\right|>\Delta E_p$, i.e. the work to be done against friction is clearly larger than what can be provided by the gravitational potential energy. So the rubber eraser cannot reach the very lowest part of the track, and (C) is correct.