Mathematics 9709/41 — May/June 2021

Walkthrough

In this problem the load is pulled up an inclined plane at constant speed. The key observation is that "constant speed" means there is no acceleration, so the resultant force on the load along the plane must be zero.

Three forces matter along the plane:

• the winch force $T$ up the plane;
• the component of the weight acting down the plane, $50g \sin 60^\circ$;
• the resistance of $100\text{ N}$, also acting down the plane.

We resolve the weight into a component parallel to the plane. The weight itself is $50g$, and its component down the slope is $50g \sin 60^\circ$. With $g = 10\text{ m s}^{-2}$, this is

$$500 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \approx 433\text{ N}$$

Because the acceleration is zero, the upward winch force must equal the sum of the downward component and the resistance:

$$T = 50g \sin 60^\circ + 100 \approx 533\text{ N}$$

Work done by a constant force is force times displacement in the direction of the force. The winch pulls through $5\text{ m}$ in its own direction, so

$$W = T \times 5 \approx 2670\text{ J}$$

This is the total work done by the winch: it lifts the load, increasing its potential energy, and also does work against the resistance.

Key Takeaways

The question tests the relationship between constant velocity and dynamic equilibrium. When a body moves at constant speed in a straight line, the resultant force on it is zero. This lets us find the unknown driving force before calculating work. It also tests resolving a weight into components on an inclined plane and using

$$W = Fd$$

for a force acting in the direction of motion.

Common Mistakes

• Using the whole weight $50g$ instead of its component down the plane $50g \sin 60^\circ$.
• Forgetting to add the $100\text{ N}$ resistance to the force that the winch must exert.
• Setting the winch force equal to the weight or to the resistance alone.
• Omitting the method: the mark scheme awards marks for resolving forces along the plane and for using $\text{work} = \text{force} \times \text{distance}$.
• Using the vertical height instead of the distance along the plane when using the work formula directly.

Things to Be Careful About

• State clearly that constant speed means zero acceleration; this justifies the force balance.
• Keep the direction of each force correct: the winch force acts up the plane, while gravity's component and the resistance act down the plane.
• Work is only $Fd$ when the force and displacement have the same direction; here the winch pulls along the plane, so the angle between them is $0^\circ$.
• Use $g = 10\text{ m s}^{-2}$, and round appropriately to give the final answer as $2670\text{ J}$.