WAEC Physics 2023 Paper

Concept:The forces opposing upward motion on an inclined plane are the component of weight acting down the plane and the frictional force.Explanation:For a body on an incline, the normal reaction is $N = mg\cos\theta$.Friction opposing motion is $f = \mu N = \mu mg\cos\theta$.The weight component pulling the device down the plane is $mg\sin\theta$.Total opposing force is $mg\sin\theta + \mu mg\cos\theta$.So, total opposing force $= mg(\sin\theta + \mu\cos\theta)$.Substitute $m = 100\,kg$, $g = 10\,\mathrm{m\,s^{-2}}$, $\theta = 30^\circ$, and $\mu = 0.25$:$= 100 \times 10 (\sin 30^\circ + 0.25\cos 30^\circ)$$= 1000(0.5 + 0.25 \times 0.8660)$$= 1000(0.5 + 0.2165) = 1000 \times 0.7165$$= 716.5\,\mathrm{N}$Answer:Total opposing force $= 716.5\,\mathrm{N}$. Therefore, the correct option is D.