Incorrect exam question? A package moving up an inclined plane, find range of values of coefficient of friction

Question

The following question was in a recent exam:

A package P of weight $10$ N is moving up an inclined plane under the action of a horizontal force of magnitude $20$ N. The plane is inclined at angle $\alpha$ to the horizontal, where $\tan{\alpha}=\frac{3}{4}$. Package P is modelled as a particle. Find the range of values $\mu$, the coefficient of friction.

To answer this I found $R=20$N and the force acting up the plane which is $10$ N. From here I feel like there is no way to continue as I do not know acceleration.

The given answer is $\mu \leq \frac{1}{2} $, as they use $20\mu \leq 10. $

This assumes acceleration is either $0$ or positive. I argue there is nothing in the question that eliminates the possibility that acceleration could be negative and that the particle is slowing down, in which case $\mu$ could be greater than $\frac{1}{2}$.

Can someone confirm this, or explain why I am wrong? Many thanks.

Answer 1

The force perpendicular to the supporting plane is $10\cos\alpha+20\sin\alpha$, therefore frictional force is $\mu(10\cos\alpha+20\sin\alpha)$ acting downhill. Force acting uphill is $20\cos\alpha-10\sin\alpha$. If you want the package to move upwards: $20\cos\alpha-10\sin\alpha-\mu(10\cos\alpha+20\sin\alpha)>0~\implies~\frac{1}{2}>\mu$.

Answer 2

Supposedly in this case it is to be assumed that the force pushed the particle from rest despite the fact that in the ideal conditions outlined the force would not be sufficient to overcome friction if $\mu=\frac{1}{2}$. So then we can assume acceleration is 0 or positive in the direction of movement, thus getting the desired answer.

Applied math. Yuck.