A 20 kg block is placed on a rough surface inclined at 30° to the horizontal - NSC Physical Sciences - Question 2 - 2021 - Paper 1

The free-body diagram should include the following forces acting on the block:

• Weight (W) acting downward
• Normal force (N) acting perpendicular to the surface
• Frictional force (f_k) acting opposite to the direction of motion
• Applied force (F) parallel to the incline.

To find the magnitude of the force F, we can apply Newton's Second Law. Since the block moves with constant velocity, the net force is zero.

egin{align*} F - f_k - W imes ext{sin}(30^ ext{o}) &= 0
W &= mg = 20 ext{kg} imes 9.81 ext{m/s}^2 \ W &= 196.2 ext{N}
F - 18 ext{N} - 196.2 imes ext{sin}(30^ ext{o}) &= 0
F - 18 ext{N} - 98.1 ext{N} &= 0
F &= 116.1 ext{N} ext{Thus, the magnitude of force } F ext{ is approximately } 116 ext{ N.}
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When the force F is removed, the net force acting on the block moving from point X to point Y can be calculated as:

egin{align*} ext{Net Force} = -f_k - W imes ext{sin}(30^ ext{o})\ ext{Net Force} = -18 ext{N} - 98.1 ext{N} = -116.1 ext{N} herefore ext{Net Force is } -116.1 ext{ N (acting down the incline).} ext{Note: The negative sign indicates the force is acting opposite to the direction of motion.} ext{The net force causing deceleration until it stops at point Y.}

Using the kinematic equation: egin{align*} v_f^2 = v_i^2 + 2a d
0 = (2 ext{ m/s})^2 + 2(-5.8 ext{m/s}^2)d
0 = 4 - 11.6d
11.6d = 4
d = rac{4}{11.6}
d ext{ is approximately } 0.34 ext{ m.} herefore ext{The distance between points X and Y is } 0.34 ext{ m.}