Mathematics 9709/43 — October/November 2016

Walkthrough

Start by noting that the block is raised at constant speed, so there is no acceleration. This tells us that the upward force from the crane exactly equals the downward forces: the block's weight $mg$ and the constant resistance $25\text{ N}$.

The work done by the crane is the energy it transfers to the block and its surroundings. Part of that energy increases the block's gravitational potential energy, and part is used to overcome the resistance.

First calculate the gain in gravitational potential energy:

$$\Delta \text{PE} = mgh = 50g \times 3.5.$$

With $g = 10\text{ m s}^{-2}$ this gives $1750\text{ J}$. This is the B1 mark in the mark scheme.

Next calculate the work done against the constant resistance. Since resistance is a constant force of $25\text{ N}$ acting over a distance of $3.5\text{ m}$, this work is

$$25 \times 3.5 = 87.5\text{ J}.$$

Add the two energy terms to find the total work done by the crane:

$$1750 + 87.5 = 1837.5\text{ J}.$$

The method of adding the PE gain and the work against resistance is what the mark scheme requires for the M1 mark; the final value gains A1.

Key Takeaways

• Work done by a crane is not simply $mgh$ when a resistance acts; the extra work needed to overcome the resistance must be included.
• At constant speed, the net change in kinetic energy is zero, so the work done by the lifting force equals the gain in gravitational potential energy plus the work done against resistance.
• The work done by a constant force is force multiplied by distance moved in the direction of the force.

Common Mistakes

• Forgetting to include the work done against the resistance and giving only $1750\text{ J}$.
• Trying to find the work done using the net force. Since the block moves at constant speed, the net force is zero; the crane's force is the weight plus the resistance, not the net force.
• Using $g = 9.8$ when the syllabus or question has specified $g = 10$; this changes the numerical answer.

Things to Be Careful About

• Use SI units: mass in kg, distance in m, force in N, work in J.
• State which value of $g$ you are using. Here $g = 10\text{ m s}^{-2}$ gives the mark scheme values.
• The mark scheme accepts $1837.5\text{ J}$ or $1840\text{ J}$ (3 s.f.).
• The resistance does negative work on the block, but the crane must do positive work to overcome it, so the two contributions are added for the crane's work.

Walkthrough

Power measures how quickly energy is transferred or work is done. The crane does a total of $1837.5\text{ J}$ of work in raising the block, and this takes $2$ seconds. Therefore the average power is

$$P = \frac{W}{t} = \frac{1837.5}{2} = 918.75\text{ W}.$$

This is the core M1/A1 step: use $P = W/t$ and substitute correctly.

An alternative route is to use $P = Fv$. At constant speed the distance $3.5\text{ m}$ is covered in $2\text{ s}$, so

$$v = \frac{3.5}{2} = 1.75\text{ m s}^{-1}.$$

The upward force exerted by the crane still has to balance the weight and the resistance:

$$F = 50g + 25 = 525\text{ N}.$$

Thus

$$P = Fv = 525 \times 1.75 = 918.75\text{ W}\approx 919\text{ W}.$$

Both methods give the same result. To three significant figures, the power is $919\text{ W}$.

Key Takeaways

• Power is the rate of doing work, $P = W/t$.
• For constant force and speed, power may also be calculated using $P = Fv$.
• The force in $P = Fv$ must be the force exerted by the crane, not the net force and not just the resistance.

Common Mistakes

• Computing $2 \div 1837.5$ instead of $1837.5 \div 2$.
• Using only $50g \times 3.5 / 2 = 875\text{ W}$, forgetting the resistance.
• Using $P = Fv$ with $F = 25\text{ N}$ (the resistance only) and forgetting the weight.

Things to Be Careful About

• Keep the working in SI units: work in J, time in s, power in W.
• The exact value is $918.75\text{ W}$; give the final answer as $919\text{ W}$ to 3 significant figures, which is what the mark scheme expects.
• If you use $P = Fv$, remember that $v$ must be the speed in the direction of the force and the block is moving at constant speed.
• Round only at the final stage to avoid accumulation of rounding errors.