Mathematics 9709/42 — February/March 2023

Walkthrough

The crate moves at constant speed, which tells us the acceleration is zero. By Newton's first law, zero acceleration means no resultant force, so the pulling force from the rope must exactly match the $600\text{ N}$ resistance. This is the key first step: the winch is effectively pulling with a $600\text{ N}$ force.

Next, the winch does work at $4.5\text{ kW}$. Convert this to watts: $4500\text{ W}$. Power measures how quickly work is done, and for a constant force moving an object at constant speed, $P = Fv$. Substituting $P = 4500$ and $F = 600$ gives $v = 7.5\text{ m s}^{-1}$ — the crate travels at $7.5$ metres per second.

Finally, time is distance divided by speed: $t = 15/7.5 = 2$ seconds. The crate takes 2 seconds to cover the 15 metres.

A useful alternative viewpoint (accepted by the mark scheme) is to use $P = \Delta W/\Delta t$ directly: the work done is force times distance, $600 \times 15$, so $4500 = 600 \times 15/t$, giving $t = 2$ the same way.

Key Takeaways

• Constant speed means equilibrium: zero acceleration implies zero net force, so the driving force equals the resistance.
• $P = Fv$: power is the rate of doing work; for a constant force and constant speed it is force times velocity.
• Time from constant motion: with constant speed, $t = d/v$.

Common Mistakes

• Substituting the crate's mass $200\text{ kg}$ into part (a) — mass is unnecessary because the speed is constant, so there is no acceleration and no $F = ma$ step.
• Forgetting to convert $4.5\text{ kW}$ to $4500\text{ W}$ before using $P = Fv$.
• Treating the winch power as if it were a force — power is not a force; it is force times velocity.

Things to Be Careful About

• Units: power must be in watts (4500, not 4.5) before substitution.
• Constant speed is the key condition: it converts the winch power into $Fv$ with $F = 600\text{ N}$.
• Keep the driving force and resistance conceptually separate: at constant speed they are equal in magnitude, but they act on the crate from opposite sides.

Walkthrough

Once the rope snaps, the winch no longer applies any force, so the crate is no longer pulled. The only horizontal force acting on it is the $600\text{ N}$ resistance, which now opposes the motion and slows the crate down. This is why Newton's second law here contains exactly two terms — the mass times acceleration on one side and the single resistance force on the other; the mark scheme highlights "2 terms only" to reinforce that the driving force has disappeared.

Taking the direction of motion as positive, the resistance is negative: $-600 = 200a$, so $a = -3\text{ m s}^{-2}$. The negative sign simply means the crate decelerates. Because the resistance is constant, the force — and therefore the acceleration — is constant, so the suvat equations are valid.

At the instant the rope breaks, the crate is still moving with the constant speed found in part (a), $u = 7.5\text{ m s}^{-1}$. It comes to rest, so the final velocity is $v = 0$. Using $v = u + at$, we get $0 = 7.5 - 3t$, giving $t = 2.5$ seconds. That is the time from the moment the rope breaks until the crate stops.

Key Takeaways

• When a driving force is removed, the remaining forces (here just the resistance) determine the acceleration via $F = ma$.
• A constant resistance means constant deceleration, so the suvat equations apply.
• $v = u + at$ with $v = 0$ directly gives the stopping time.

Common Mistakes

• Including the winch driving force in Newton's second law after the rope breaks — that force no longer exists.
• Using the power value ($4.5\text{ kW}$) as a force in part (b) — the winch force vanishes instantaneously when the rope breaks.
• Getting the sign of the acceleration wrong, which would produce a negative time; the mark scheme requires an equation that leads to a positive $t$.

Things to Be Careful About

• Sign convention: with the direction of motion taken as positive, the resistance must be written as $-600$, giving $a = -3\text{ m s}^{-2}$.
• The initial speed is $7.5\text{ m s}^{-1}$ (carried from part (a)), not the distance $15\text{ m}$.
• The stopping time is measured from the moment the rope breaks — the $2$ s from part (a) is not part of this interval.
• The mass matters here: $a = F/m = 600/200 = 3\text{ m s}^{-2}$, so do not drop the $200\text{ kg}$.