JAMB Physics 2018 Paper

Concept:Efficiency tells us how much of the work input is usefully used to lift the load.For an inclined plane, efficiency relates mechanical advantage to velocity ratio.Explanation:For an inclined plane at angle $\theta$ to the horizontal:$V.R = \frac{1}{\sin\theta}$Given $\theta = 30^\circ$:$V.R = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2$The load to be moved is the weight of the mass:$\text{Load} = mg = 20 \times 10 = 200\,N$Efficiency is given by:$\text{Efficiency} = \frac{M.A}{V.R} \times 100\%$Substitute the given efficiency:$\frac{75}{100} = \frac{M.A}{2}$$M.A = \frac{75 \times 2}{100} = 1.5$Also, mechanical advantage is:$M.A = \frac{\text{Load}}{\text{Force}}$Therefore:$1.5 = \frac{200}{\text{Force}}$$\text{Force} = \frac{200}{1.5} = 133.3\,N$Answer:The force required to pull the load up the plane is $133.3\,N$.The correct option is C.