A-Level Mathematics May/June 2025 Q1: A block of mass 12kg is being pulled by a rope up a rough plane. The plane is inclined…

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  Let $T$ be the tension in the rope, $R$ be the normal reaction force, and $F$ be the frictional force. The plane is inclined at an angle $\theta = 20^{\circ}$. The mass of the block is $m = 12\text{kg}$ and its acceleration is $a = 2\text{ms}^{-2}$. The coefficient of friction is $\mu = 0.4$. We use $g=10\text{ms}^{-2}$.

  1. Resolve forces perpendicular to the plane: There is no acceleration perpendicular to the plane, so the forces are in equilibrium. $R - mg \cos 20^{\circ} = 0$ $R = 12g \cos 20^{\circ}$ $R = 12(10) \cos 20^{\circ}$ $R = 120 \cos 20^{\circ} \approx 112.763\text{ N}$

  2. Apply Newton's Second Law parallel to the plane (up the slope): The net force in the direction of motion is equal to mass times acceleration ($F_{net} = ma$). The block is accelerating up the plane. Forces acting up the plane: $T$ Forces acting down the plane: Frictional force ($F$) and the component of weight ($mg \sin 20^{\circ}$)

  Equation of motion: $T - F - mg \sin 20^{\circ} = ma$

  3. Substitute for friction and solve for T: The block is moving, so the frictional force is given by $F = \mu R$. $F = 0.4 \times (120 \cos 20^{\circ})$

  Substitute this and the other values into the equation of motion: $T - (0.4 \times 120 \cos 20^{\circ}) - 12(10) \sin 20^{\circ} = 12(2)$ $T - 48 \cos 20^{\circ} - 120 \sin 20^{\circ} = 24$ $T = 24 + 120 \sin 20^{\circ} + 48 \cos 20^{\circ}$ $T = 24 + 41.042... + 45.105...$ $T = 110.147...$

  Final Answer: The tension in the rope is $110\text{ N}$ (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly resolving forces perpendicular to the plane to find the normal reaction, R. The expression R = 12g cos 20 or the value 112.76... is sufficient.
  • M1 — Applying Newton's Second Law parallel to the plane. The equation must contain four terms: Tension (T), Friction (F), the component of weight (12g sin 20), and the resultant force (12 × 2), even if signs are incorrect.
  • DM1 — Substituting the formula for friction, F = 0.4R, using the calculated expression for R, into the Newton's Second Law equation to form an equation in T only.
  • A1 — Obtaining the correct final answer T = 110 N (or anything that rounds to 110) from fully correct working.

  Common mistakes

  • Mixing up sine and cosine when resolving the weight component, for example using R = 12g sin 20° or using 12g cos 20° as the force component parallel to the plane.
  • Assuming the normal reaction R is equal to the weight mg. This is only true on a horizontal surface; on an inclined plane, R is the component of weight perpendicular to the surface, mg cos θ.
  • Applying the frictional force in the wrong direction. Friction always opposes motion, so as the block is pulled up the plane, friction must act down the plane.
  • Omitting one of the forces when applying Newton's Second Law, most commonly forgetting the component of weight acting down the slope (12g sin 20°).

  Examiner tip: Always begin a mechanics problem involving an inclined plane by drawing a clear free-body diagram and resolving forces perpendicular and parallel to the plane.

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