OCR GCSE Combined Science
Topic questions Forces Work done and energy transfer
16 questions · 94 marks · about 2 hours
A lift is stationary 20 m above the ground. The lift moves upwards until it is now 100 m above the ground. The lift then remains stationary. The mass of the lift is 450 kg.
What is the increase in gravitational potential energy of the lift?
Gravitational field strength = 10 N / kg
Use the equation: gravitational potential energy = mass × gravitational field strength × height
A 45.0 J
B 56.3 J
C 360 000 J
D 450 000 J
Two students do an experiment to measure their power by walking up steps.
This is the data from student P:
• Weight = 600 N
• Number of steps = 250
• Height of each step = 0.20 m
Calculate the work done by student P.
This is the data from student Q:
• Work done = 36 000 J
• Time taken = 240 s
Calculate the power of student Q.
Give your answer in kW.
Describe how the students carry out this experiment. Include any equipment they use.
Student P carries out the experiment five more times and calculates their power.
Here are student P ’s results.
| Try 1 | Try 2 | Try 3 | Try 4 | Try 5 | |
| Power (W) | 120 | 121 | 122 | 121 | 123 |
How can student P ’s results be described?
Tick one (✓) box in each row.
Explain your answer to (d)(i).
Which two units measure the same quantity?
A W and J
B N m and J
C kg and N
D V and A
Two students do an experiment to measure their power by walking up steps.
This is the data from student P:
• Weight = 600 N
• Number of steps = 250
• Height of each step = 0.20 m
Calculate the work done by student P.
This is the data from student Q:
• Work done = 36 000 J
• Time taken = 240 s
Calculate the power of student Q.
Give your answer in kW.
Describe how the students carry out this experiment. Include any equipment they need to use.
Student P carries out the experiment five more times and calculates their power.
Here are student P ’s results.
| Try 1 | Try 2 | Try 3 | Try 4 | Try 5 | |
| Power (W) | 120 | 121 | 122 | 121 | 123 |
How can student P ’s results be described?
Tick one (✓) box in each row.
Explain your answer to (d)(i).
A car has a mass of 1100 kg.
Calculate the kinetic energy of the car when it is travelling at 10 m / s. Use the equation: kinetic energy = 
× mass × (speed)2
When the car has a kinetic energy of 140 000 J, the driver presses the brakes.
Calculate the braking force if the car stops in a distance of 40 m.
Use the equation: work done = force × distance
A child is holding a marble at a height of 0.8 m.
The mass of the marble is 0.015 kg.
Calculate the gravitational potential energy of the marble.
Gravitational field strength = 10 N / kg.
Use the equation: gravitational potential energy = mass × gravitational field strength × height
The child rolls the marble along a track. The diagram shows the marble and track.
The kinetic energy of the marble at the bottom of the track is less than the potential energy of the marble calculated at the top of the track.
Suggest and explain why.
The mass of the marble is 0.015 kg.
The kinetic energy of the marble at the bottom of the track is 0.03 J.
Calculate the speed of the marble at the bottom of the track. Use the equation: kinetic energy = 
× mass × (speed)2
Speed of marble =........ m / s
The child now has a small marble and a large marble. The child rolls the small marble across the floor. The small marble collides with the stationary large marble.
The diagram shows the marbles before and after the collision.
Before the collision
After the collision
The mass of the small marble is 5 grams. The mass of the large marble is 20 grams.
Use the information in the diagram to calculate the speed of the large marble after the collision.
Use the Equation Sheet.
Speed of large marble =........ m / s
A teacher lifts a weight of 45 N through a height of 1.8 m.
Calculate the work done by the teacher.
Use the equation: work done = force × distance
Another teacher does 1500 J of work.
They take 5 seconds to do the work.
Calculate their power.
Give the unit in your answer.
Use the equation: power = work done / time
The braking distance for a car is 25 m. The braking force is 4000 N.
Calculate the work done by the brakes.
Use the equation: work done = force × distance
A 1.6 J
B 100 J
C 160 J
D 100 000 J
Explain in words what is meant by power.
A weightlifter lifts a 200 pound (200 lb) mass.
1 lb = 0.454 kg.
Calculate the weight of the mass in Newtons.
Use the equation: gravitational force = mass × gravitational field strength
Gravitational field strength = 10 N / kg.
This is some data for a weightlifter.
Weight lifted = 620 N Height of the lift = 1.6 m Time taken to complete lift = 4.0 s
Calculate the power generated by the weightlifter.
Use the equations: work done = force × distance power = work done / time
A man weighs 700 N and climbs a staircase 5 m high.
How much work does he do?
Use the equation: work done = force × distance
A 140 J
B 350 J
C 695 J
D 3500 J
A student measures the acceleration of a trolley on a horizontal surface.
Fig. 15.1 is a diagram of her experiment.
The weight of the masses accelerates the trolley.
Define the term weight.
Show that the weight of a 100 g mass is 1 N.
Friction can change the results of this experiment.
Suggest one way friction can be reduced.
The tall parts of the card on the trolley (U and V) pass through the light gate, as shown in Fig. 15.2.
The velocity of U is 0.5 m / s.
The velocity of V is 1.0 m / s.
Calculate the acceleration of the trolley.
Use the equation: (final velocity)2 – (initial velocity)2 = 2 
acceleration distance
Acceleration =........ m / s2
When the trolley reaches the edge of the surface, it is moving at 1.2 m / s.
The mass of the trolley is 1.0 kg.
Calculate the kinetic energy of the trolley.
The student repeats the experiment with different masses.
This is a graph showing her results.
The student says, ‘The graph shows that the acceleration of the trolley is directly proportional to the force.’
Is the student correct?
Explain your answer.
A student completes four calculations for work done.
Which row of the table is correct?
Use the equation: work done = force × distance
| Force (N) | Distance (m) | Work done (J) | |
| A | 100 | 2 | 200 |
| B | 100 | 3 | 33 |
| C | 200 | 2 | 100 |
| D | 200 | 3 | 467 |
A student uses a small motor to lift a toy car through a vertical distance of 1.0 m.
The car has a weight of 0.05 N.
Calculate the work done when lifting the car through this distance.
Calculate the power of the motor if the car takes 5.0 seconds to travel the 1.0 m distance.
State the unit.
The motor uses 2 cells in series.
Each cell has a potential difference of 1.5 V.
Write down the total potential difference of the cells.
The motor has a resistance of 6.0 
.
Calculate the current in the circuit when the motor is in use.
Use the equation: Potential difference = Current × Resistance
The student wishes to increase the time taken to lift this toy car vertically through the 1.0 m distance shown.
Suggest a change he could make to this experiment to achieve this.
A student lifts four different objects onto a set of shelves.
| Object | Mass (kg) | Height lifted (m) |
| A | 0.1 | 2.2 |
| B | 0.3 | 1.5 |
| C | 0.4 | 1.7 |
| D | 0.5 | 2.0 |
Which object gains the most gravitational potential energy?
A student uses a small motor to lift a toy car through a vertical distance of 1.0 m.
The car has a weight of 0.05 N.
Calculate the work done when lifting the car through this distance.
Calculate the power of the motor if the car takes 5.0 seconds to travel the 1.0 m distance.
State the unit.
The motor uses 2 cells in series.
Each cell has a potential difference of 1.5 V.
Write down the total potential difference of the cells.
The motor has a resistance of 6.0 .
Calculate the current in the circuit when the motor is in use.
Use the equation: Potential difference = Current × Resistance
The student wishes to increase the time taken to lift this toy car vertically through the 1.0 m distance shown.
Suggest a change he could make to this experiment to achieve this.
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