Mathematics 9709/42 — February/March 2022

Cambridge AS Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Energy, Work and Power · Kinematics of Motion in a Straight Line · Forces and Equilibrium · Newton's Laws of Motion · Momentum

Q1Energy, Work and PowerFree sample

A crane is used to raise a block of mass $600\text{ kg}$ vertically upwards at a constant speed through a height of $15\text{ m}$ . There is a resistance to the motion of the block, which the crane does $10\,000\text{ J}$ of work to overcome.

(a)

Find the total work done by the crane.

DifficultyMedium-Easy

Worked solution

Approach

Since the block is raised at constant speed, its kinetic energy does not change. The work done by the crane therefore equals the gain in gravitational potential energy plus the work done against the resistance.

Working

The gain in gravitational potential energy is

600 \times 10 \times 15 = 90\,000\ \text{J}

The work done against the resistance is given as $10\,000\ \text{J}$ . Hence the total work done by the crane is

90\,000 + 10\,000 = 100\,000\ \text{J}

Answer

100\,000\ \text{J}

Final answer

100000 J

Detailed explanation

Walkthrough

Start by recognizing that the block moves at constant speed vertically. Constant speed means zero acceleration, so there is no change in kinetic energy; all the work done by the crane goes into raising the block's potential energy and overcoming the resistance.

First calculate the gain in gravitational potential energy using $\Delta E_{p} = mgh$ . With $m = 600\,\text{kg}$ , $g = 10\,\text{m s}^{-2}$ and $h = 15\,\text{m}$ , this gives $90\,000\ \text{J}$ .

The resistance does not store energy: the crane must supply $10\,000\ \text{J}$ purely to overcome it. Therefore the total work done by the crane is the sum of these two amounts, $90\,000 + 10\,000 = 100\,000\ \text{J}$ .

Key Takeaways

This question combines gravitational potential energy with work done against a resistance. When an object moves at constant speed, the net work done on it is zero, so the work done by the driving force must balance both the gravitational work and any resistive work.

Common Mistakes

Forgetting to add the $10\,000\ \text{J}$ of work done against the resistance.
Using the mass times height without multiplying by $g$ .
Assuming there is a change in kinetic energy when the speed is constant.

Things to Be Careful About

Use a consistent value of $g$ ; the mark scheme here uses $g = 10\,\text{m s}^{-2}$ , giving $600 \times 10 \times 15 = 90\,000$ .
The work against resistance is given in joules, so no conversion is needed before adding.
The result is a scalar quantity measured in joules.

Techniques used

calculate gravitational potential energy gainedapply work-energy principle at constant speedadd work done against resistance

(b)

Given that the average power exerted by the crane is $12.5\text{ kW}$ , find the total time for which the block is in motion.

DifficultyMedium-Easy

Worked solution

Approach

Use the relationship that the average power is the total work done divided by the time taken, i.e. $P = \frac{W}{t}$ , so $W = Pt$ . Substitute the total work done from part (a) and the power in watts, then solve for $t$ .

Working

Convert the average power to watts:

P = 12.5\ \text{kW} = 12\,500\ \text{W}

Use $\text{work done} = \text{power} \times \text{time}$ :

100\,000 = 12\,500 \, t

Solve for $t$ :

t = \frac{100\,000}{12\,500} = 8\ \text{s}

Answer

8\ \text{s}

Final answer

8 s

Detailed explanation

Walkthrough

The average power exerted by the crane is $12.5\,\text{kW}$ . Power is the rate at which work is done, so $P = \frac{W}{t}$ , equivalently $W = Pt$ .

Before substituting, convert the power into watts because the work done in part (a) is measured in joules and time will be in seconds: $12.5\,\text{kW} = 12\,500\,\text{W}$ .

Using the total work done by the crane from part (a), $W = 100\,000\,\text{J}$ , the equation becomes $100\,000 = 12\,500\, t$ . Dividing both sides by $12\,500$ gives $t = 8$ seconds.

Key Takeaways

Power is work done per unit time. This question shows how to rearrange $P = \frac{W}{t}$ to find time once total work and average power are known.

Common Mistakes

Forgetting to convert $12.5\,\text{kW}$ into $12\,500\,\text{W}$ .
Using the potential energy gain $90\,000\,\text{J}$ instead of the total work done $100\,000\,\text{J}$ .
Reversing the fraction, e.g. calculating $\frac{12\,500}{100\,000}$ .

Things to Be Careful About

The mark scheme follows on from the candidate's total work done in part (a); here that is $100\,000\,\text{J}$ .
Units must be consistent: joules, watts and seconds.
The final time must be stated with seconds.

Techniques used

convert average power to wattsapply work equals power times timesolve for time

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line5M
Q3Energy, Work and Power · Forces and Equilibrium · Newton's Laws of Motion5M
Q4Energy, Work and Power · Newton's Laws of Motion6M
Q5Forces and Equilibrium7M
Q6Kinematics of Motion in a Straight Line11M
Q7Forces and Equilibrium · Kinematics of Motion in a Straight Line · Newton's Laws of Motion · Momentum12M

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