# Motion on a slope - A Level

A particle P of mass 5kg and particle Q of mass 1kg are connected by a light inextensible string. P lies on a slope inclined at a 60 degree angle to the horizontal. The string passes from P parallel to the line of greatest slope, and runs over a light pulley that can freely rotate. Then descends vertically to Q. The coefficient of friction between P and the slope is 0.7. Find the force of friction acting on P and it’s direction and find the acceleration of P when the system is released from rest.

I was able to find the frictional force 17.15N up the slope) but I can’t find the acceleration. I have tried to use the equation 5gsin(60)-17.15-9.8 = 5a with 9.8 being the weight of Q, but I get the answer of 3.1m/s/s for a, bit the mark scheme says that a (acceleration) is 2.58m/s/s but I don’t see how. Could you please give me some advice on how to get 2.58?

You have done all correct, you just messed up in the calculation. The equation $$5g\sin60° -17.15-9.8=6a$$ has the solution $$\boxed{a=2.58\,\text{m}\,\text{s}^{-2}}$$ as required.
Remember the mass is $$5\,\text{kg}$$ for $$\text{P}$$ and $$1\,\text{kg}$$ for $$\text{Q}$$, so $$5g\sin60°-17.15-9.8 = 6a$$
This will give you $$a = 2.58\,\text{m}/\text{s}^2$$.