A toy car P (mass 4 kg) rests on a rough horizontal plane, with coefficient of friction $\frac12$ between P and the plane, and is connected by a light inextensible string over a smooth pulley X to toy car Q (mass 8 kg) held on a smooth plane inclined at 30° to the horizontal. When Q is released, find the acceleration of the system and the tension in the string, in terms of $g$.

For Q, the driving force down the incline is $8g\sin30°=4g$; the resisting force on P is friction $\mu\times4g=\frac12\times4g=2g$. The net force on the system is $4g-2g=2g$ and the total mass is $4+8=12$ kg, so $a=\frac{2g}{12}=\frac{g}{6}$. For P: $T-2g=4a=4\left(\frac{g}{6}\right)=\frac{2g}{3}$, so $T=2g+\frac{2g}{3}=\frac{8g}{3}$. Using Q's full weight $8g$ as the driving force instead of the component along the incline, $8g\sin30°=4g$, gives $a=\frac{g}{2}$, $T=4g$. Computing the friction force using Q's weight instead of P's own weight (which is what actually determines the normal reaction under P) gives a driving force that exactly cancels the (wrong) friction, so $a=0$, $T=4g$. Reaching the correct acceleration but then forgetting to include the friction force $2g$ when writing Newton's second law for P gives $T=4a=\frac{2g}{3}$.