The fig above shows a block of mass m sliding down a rough inclined plane QP at angle \( \theta \) . The forces acting on the block along QP are
Attached Figure / Diagram
A
mg sinθ and the normal reaction
B
mg sinθ and the force of friction
C
mg sinθ and the normal reaction
D
mg cosθ and the normal reaction
Verified Correct Answer: B.
Physical Principles & Numerical Solution
The weight mg acts vertically downward.
Along the incline (parallel to QP):
The component of weight down the plane: mg sin\theta
Since the plane is rough and the block is sliding down, kinetic friction acts up the plane, opposing the motion.
The normal reaction (N) acts perpendicular to the plane (not along QP).
So, forces along QP (parallel to the incline) are:
1. mg sin\theta (down the plane)
2. Force of friction (up the plane)
Source: Myschool Past Questions Archive (Record #41734)