Work

1. Overview

Work is the measure of energy transfer that occurs when an object is moved over a distance by an external force. Understanding work is essential because it links the concept of forces to energy changes, allowing us to quantify how much energy is needed to perform physical tasks.

Key Definitions

• Work Done: The product of the force applied to an object and the distance moved in the direction of that force.
• Joule (J): The SI unit of work and energy. One joule of work is done when a force of one newton moves an object one meter.
• Energy Transferred: The amount of energy that changes from one form to another (or moves from one object to another) during a process.

Core Content

Work as Energy Transfer

In physics, "doing work" is the same as "transferring energy." If you do 50 J of work to lift a box, you have transferred 50 J of energy from your chemical store to the box’s gravitational potential energy store.

• Mechanical Work: Energy transferred by a force moving an object (e.g., pushing a car).
• Electrical Work: Energy transferred by a current (e.g., a battery powering a motor).

Calculating Work Done

To calculate the mechanical work done, we use the force applied and the distance moved. It is important that the distance used is specifically the distance moved in the direction of the force.

Worked Example: A crate is pushed with a constant force of 150 N across a warehouse floor for a distance of 10 m. Calculate the work done.

• $W = F \times d$
• $W = 150 \text{ N} \times 10 \text{ m}$
• $W = 1500 \text{ J}$ (or $1.5 \text{ kJ}$)

Extended Content (Extended Only)

There are no additional Supplement objectives for this specific sub-topic.

Key Equations

The Work Equation: $$W = F \times d$$

• $W$ = Work done (Joules, J)
• $F$ = Force (Newtons, N)
• $d$ = Distance moved in the direction of the force (metres, m)

Note: Since Work Done = Energy Transferred, you may also see this written as $\Delta E = F \times d$.

Common Mistakes to Avoid

• ❌ Wrong: Thinking that holding a heavy weight stationary is "doing work."
  • ✓ Right: Even if it feels tiring, if there is no distance moved, the work done is zero.
• ❌ Wrong: Forgetting to convert units (e.g., using cm instead of m).
  • ✓ Right: Always convert distance to metres and force to Newtons before calculating.
• ❌ Wrong: Squaring the distance or velocity as if you were calculating Kinetic Energy.
  • ✓ Right: The formula is simply $F \times d$. Do not square any numbers in the work equation.
• ❌ Wrong: Confusing Power with Work.
  • ✓ Right: Work is the total energy transferred ($J$); Power is the rate ($J/s$) at which that work is done.
• ❌ Wrong: Focusing only on the force applied without considering the distance.
  • ✓ Right: A large force over zero distance results in zero work; always look for the product of both.

Exam Tips

1. Check the Units: Examiners often give the distance in centimeters (cm) or the force in kilonewtons (kN). Convert these to meters (m) and Newtons (N) immediately to avoid losing easy marks.
2. Direction Matters: If an object moves at a right angle to the force (for example, carrying a box horizontally while the lifting force is vertical), no work is being done by that specific force on the object.
3. The "Work-Energy" Link: If a question asks for "energy transferred" and gives you force and distance, use the work formula! Students often get stuck looking for an energy formula when the work formula is the one required.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student lifts a book of mass 1.5 kg from the floor to a shelf 1.2 m above the floor.

(a) Define work done. [2]

(b) Calculate the work done by the student in lifting the book to the shelf. Assume the acceleration due to gravity, $g$, is 9.8 N/kg. [3]

Worked Solution:

(a)

1. Work done is the energy transferred when a force causes a displacement. Defining work done in terms of energy transfer and displacement.

How to earn full marks:

• State that work done is related to energy transfer (1 mark)
• Mention displacement or distance moved in the direction of the force (1 mark)

(b)

1. Calculate the force needed to lift the book, which is equal to its weight. $F = mg = 1.5 \text{ kg} \times 9.8 \text{ N/kg} = 14.7 \text{ N}$ Calculating the force (weight) needed to lift the book.

2. Calculate the work done using the formula $W = Fd$. $W = 14.7 \text{ N} \times 1.2 \text{ m} = 17.64 \text{ J}$ Calculating work done using force and distance.

3. Round the answer to 2 or 3 significant figures. $W = \boxed{17.6 \text{ J}}$ Stating the final answer with correct units.

How to earn full marks:

• Correctly calculate the force (weight) using $F=mg$ (1 mark)
• Use the correct formula for work done, $W=Fd$ (1 mark)
• Correct final answer with correct units (1 mark)

Common Pitfall: Remember that work is only done when there is a displacement. Holding the book stationary, even though it requires effort, does not constitute work done in the physics sense. Also, be sure to use the weight of the book (mg) as the force, not just its mass.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

An electric motor lifts a mass of 50 kg through a vertical height of 8.0 m in 20 s at a constant speed.

(a) Calculate the work done by the motor in lifting the mass. Assume the acceleration due to gravity, $g$, is 10 N/kg. [3]

(b) Calculate the power output of the motor. [3]

Worked Solution:

(a)

1. Calculate the force needed to lift the mass, which is equal to its weight. $F = mg = 50 \text{ kg} \times 10 \text{ N/kg} = 500 \text{ N}$ Calculating the force (weight) needed to lift the mass.

2. Calculate the work done using the formula $W = Fd$. $W = 500 \text{ N} \times 8.0 \text{ m} = 4000 \text{ J}$ Calculating work done using force and distance.

3. Stating the final answer with correct units. $W = \boxed{4000 \text{ J}}$

How to earn full marks:

• Correctly calculate the force (weight) using $F=mg$ (1 mark)
• Use the correct formula for work done, $W=Fd$ (1 mark)
• Correct final answer with correct units (1 mark)

(b)

1. Calculate the power using the formula $P = \frac{W}{t}$. $P = \frac{4000 \text{ J}}{20 \text{ s}} = 200 \text{ W}$ Calculating power using work done and time.

2. Stating the final answer with correct units. $P = \boxed{200 \text{ W}}$

How to earn full marks:

• Use the correct formula for power, $P = \frac{W}{t}$ (1 mark)
• Correct final answer (1 mark)
• Correct units (1 mark). ECF allowed from part (a).

Common Pitfall: Students often forget that the force required to lift an object is equal to its weight (mg). Also, remember that power is the rate at which work is done, so you need to divide the work done by the time taken.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A cyclist travels along a horizontal road. The cyclist exerts a constant forward force of 60 N.

(a) The cyclist travels 150 m in 25 s. Calculate the work done by the cyclist. [2]

(b) Calculate the power developed by the cyclist. [2]

(c) After travelling 150 m, the cyclist starts to cycle up a hill. Explain why the power developed by the cyclist may increase even if the forward force and speed remain constant. [4]

Worked Solution:

(a)

1. Calculate the work done using the formula $W = Fd$. $W = 60 \text{ N} \times 150 \text{ m} = 9000 \text{ J}$ Calculating work done using force and distance.

2. Stating the final answer with correct units. $W = \boxed{9000 \text{ J}}$

How to earn full marks:

• Use the correct formula for work done, $W=Fd$ (1 mark)
• Correct final answer with correct units (1 mark)

(b)

1. Calculate the power using the formula $P = \frac{W}{t}$. $P = \frac{9000 \text{ J}}{25 \text{ s}} = 360 \text{ W}$ Calculating power using work done and time.

2. Stating the final answer with correct units. $P = \boxed{360 \text{ W}}$

How to earn full marks:

• Use the correct formula for power, $P = \frac{W}{t}$ (1 mark)
• Correct final answer with correct units (1 mark). ECF allowed from part (a).

(c)

1. The power will increase because the cyclist is now doing work against gravity. Stating that work is being done against gravity.

2. The cyclist is increasing their gravitational potential energy (GPE). Linking work done to an increase in GPE.

3. The total work done is now greater because it includes work against gravity in addition to overcoming resistive forces. Explaining that the total work done is greater due to the additional work against gravity.

4. Since power is work done per unit time, and the work done increases while the time stays the same, the power must increase. Relating increased work done to increased power.

How to earn full marks:

• State that the cyclist is working against gravity (1 mark)
• Link this to an increase in gravitational potential energy (GPE) (1 mark)
• Explain that the total work done is greater because it includes work against gravity (1 mark)
• Relate the increased work done to an increase in power (1 mark)

Common Pitfall: Many students forget that when cycling uphill, the cyclist is doing work against gravity in addition to the work done to overcome friction and air resistance. This extra work done increases the power output, even if the forward force and speed remain constant.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A small electric car of mass 800 kg accelerates from rest to a speed of 15 m/s on a level road. Assume air resistance is negligible.

(a) Calculate the kinetic energy gained by the car. [2]

(b) Calculate the minimum work done by the motor in accelerating the car. [1]

(c) The car travels a distance of 75 m during this acceleration. Calculate the average force exerted by the motor on the car. [2]

(d) In reality, air resistance is not negligible. The actual force exerted by the motor is greater than calculated in (c). Explain why. [4]

Worked Solution:

(a)

1. Calculate the kinetic energy using the formula $KE = \frac{1}{2}mv^2$. $KE = \frac{1}{2} \times 800 \text{ kg} \times (15 \text{ m/s})^2 = 90000 \text{ J}$ Calculating kinetic energy using mass and velocity.

2. Stating the final answer with correct units. $KE = \boxed{90000 \text{ J}}$

How to earn full marks:

• Use the correct formula for kinetic energy, $KE = \frac{1}{2}mv^2$ (1 mark)
• Correct final answer with correct units (1 mark)

(b)

1. The minimum work done is equal to the kinetic energy gained. Equating work done to the change in kinetic energy. $W = 90000 \text{ J}$

2. Stating the final answer with correct units. $W = \boxed{90000 \text{ J}}$

How to earn full marks:

• State that work done is equal to the kinetic energy gained (1 mark)

(c)

1. Calculate the average force using the formula $W = Fd$, rearranging to $F = \frac{W}{d}$. Calculating force using work done and distance. $F = \frac{90000 \text{ J}}{75 \text{ m}} = 1200 \text{ N}$

2. Stating the final answer with correct units. $F = \boxed{1200 \text{ N}}$

How to earn full marks:

• Use the correct formula for force, $F = \frac{W}{d}$ (1 mark)
• Correct final answer with correct units (1 mark). ECF allowed from part (b).

(d)

1. Air resistance opposes the motion of the car. Stating that air resistance is a force opposing motion.

2. This means that the motor has to do additional work to overcome air resistance. Explaining that work is done to overcome air resistance.

3. The total work done by the motor is the sum of the work done to increase the kinetic energy of the car and the work done to overcome air resistance. Stating that the total work done is the sum of KE and work against air resistance.

4. Since the total work done is greater, the force exerted by the motor must be greater to achieve the same acceleration and distance. Relating increased work done to increased force.

How to earn full marks:

• State that air resistance opposes motion (1 mark)
• Explain that the motor does additional work to overcome air resistance (1 mark)
• State that the total work done is the sum of KE and work against air resistance (1 mark)
• Relate the increased work done to an increased force (1 mark)

Common Pitfall: A common mistake is to forget that the work done by the motor is not just to increase the kinetic energy of the car. In real-world scenarios, some of the work is also used to overcome resistive forces like air resistance. Therefore, the actual force exerted by the motor will be greater than the minimum force calculated when neglecting air resistance.