Laws of Motion Part 2

Concept:For motion without acceleration (constant velocity), the net force along the incline is zero. Friction always opposes the relative motion, so if the block is moving down the plane, friction acts upward along the incline.Explanation:The component of weight acting down the incline is $mg \sin \theta = 5 \times 10 \times \sin 30^\circ = 25\ \mathrm{N}$.Normal force is $N = mg \cos \theta = 5 \times 10 \times \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}\ \mathrm{N}$.Kinetic friction (acting upward) is $f_k = \mu N = \frac{\sqrt{3}}{2} \times 25\sqrt{3} = \frac{3}{2} \times 25 = 37.5\ \mathrm{N}$.Since $f_k > mg\sin\theta$, the block cannot slide down on its own. To move it down at constant speed, we apply an additional force $P$ downwards along the incline.For equilibrium along the incline: downward forces = upward force.$P + mg\sin\theta = f_k$$P + 25 = 37.5$$P = 37.5 - 25 = 12.5\ \mathrm{N}$.Answer:$12.5\ \mathrm{N}$ (Option B).