A particle $P$ with mass $0.2\,\text{kg}$ is released from rest at point $O$ on a rough plane inclined at $60^\circ$ to the horizontal, and moves down the line of greatest slope. The coefficient of friction between $P$ and the plane is $0.3$. A force of magnitude $0.6x\,\text{N}$ acts on $P$ in the direction $PO$, where $x\,\text{m}$ is the displacement of $P$ from $O$.
(i)[3]
Show that $v\frac{dv}{dx}=5\sqrt{3}-1.5-3x$, where $v\,\text{m s}^{-1}$ is the velocity of $P$ when its displacement from $O$ is $x\,\text{m}$.
(ii)[4]
Find the value of $x$ for which $P$ attains its maximum velocity, and calculate this maximum velocity.
(iii)[4]
Calculate the magnitude of the acceleration of $P$ immediately after it has first come to instantaneous rest.
Worked solution & mark scheme
This 11-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Applies Newton's Second Law parallel to the plane to get $0.2v\,\frac{dv}{dx} = 0.2g\sin60 - 0.3\times0.2g\cos60 - 0.6x$” …