Work done & energy transfer

Figure 6 shows a student about to launch a toy aeroplane horizontally.

The student pulls on the aeroplane to stretch the spring.

The student then releases the aeroplane and the spring gets shorter.

How does the elastic potential energy of the spring change just after the aeroplane is released?

Tick () one box.

The elastic potential energy decreases.

The elastic potential energy stays the same.

The elastic potential energy increases.

How does the kinetic energy of the aeroplane change just after the aeroplane is released?

Tick () one box.

The kinetic energy decreases.

The kinetic energy stays the same.

The kinetic energy increases.

As the aeroplane is launched, the mean force of the spring on the aeroplane is 8.0 N.

The distance moved by the aeroplane during the launch is 0.25 m.

Calculate the work done on the aeroplane.

Use the equation:

work done = force × distance

Choose the unit from the box.

After launching, the aeroplane detached from the spring and glided until it reached the ground.

Table 3 shows how the horizontal distance travelled varied with the extension of the spring when the aeroplane was released.

Complete Figure 7.

You should:

plot the data from Table 3 on Figure 7

draw a line of best fit.

Which two changes would make the aeroplane fly a greater horizontal distance?

Tick () two boxes.

Decrease the length of the wings of the aeroplane.

Increase the launch speed of the aeroplane.

Increase the mass of the aeroplane.

Launch the aeroplane vertically upwards.

Make the aeroplane more streamlined.

Figure 8 shows a student about to launch a toy aeroplane horizontally.

The student pulled on the aeroplane to stretch the spring.

The student then released the aeroplane.

After the aeroplane was released, the aeroplane accelerated until it passed the student’s hand.

Describe how the energy stores of the aeroplane and spring changed as the aeroplane accelerated.

The spring exerted a force on the aeroplane to make the aeroplane accelerate.

As the aeroplane accelerated, it moved through a distance of 40 cm.

The work done by the spring on the aeroplane was 3.0 J.

Calculate the mean force exerted by the spring on the aeroplane.

Use the Physics Equations Sheet.

The student launched the aeroplane at different angles to the horizontal.

The student then measured the horizontal distance travelled by the aeroplane.

Figure 9 shows the launch angle of the aeroplane.

After being launched, the aeroplane flew through the air and landed on the ground.

Figure 10 shows how the horizontal distance travelled varies with the launch angle.

The extension of the spring was the same in each test.

Describe how the horizontal distance travelled varies with the launch angle.

Explain how increasing the extension of the spring would affect the horizontal distance travelled by the aeroplane.

Figure 10 shows a karate expert breaking a wooden board with one hand.

When the hand hits the wooden board, the initial velocity of the hand is 7.5 m/s.

The change in momentum of the hand is 5.0 kg m/s.

The mass of the hand is 0.80 kg.

Calculate the final velocity of the hand.

Use the Physics Equations Sheet.

As the hand exerts a force on the wooden board, the wooden board exerts a force on the hand.

Explain how Newton’s third law applies to the hand hitting the wooden board.

Figure 11 shows the hand hitting the wooden board. When the hand hits the wooden board, the wooden board bends.

When the hand hits the wooden board, the hand moves through a distance of 1.2 cm while exerting a force on the wooden board.

The work done by the hand is 6.0 J.

Calculate the force the hand exerts on the wooden board.

Use the Physics Equations Sheet.

Figure 1 shows a boat pulling a person parasailing.

A rope attaches the person to the boat.

What type of force is the weight?

Tick (✓) one box.

Contact force

Magnetic force

Non-contact force

The mass of the person is 75 kg.

Calculate the weight of the person.

gravitational field strength = 9.8 N/kg

Use the equation:

weight = mass × gravitational field strength

The resultant force acting on the person is zero.

Which of the following describes the motion of the person?

Tick (✓) one box.

Velocity decreasing

Moving at constant velocity

Velocity increasing

The horizontal force on the person is 4300 N.

Calculate the work done by this force in moving the person a horizontal distance of 500 m.

Use the equation:

work done = force × distance

Choose the unit from the box.

The speed of the boat changes.

The height of the person above the water decreases by 18 m.

Calculate the decrease in gravitational potential energy of the person.

mass of person = 75 kg

gravitational field strength = 9.8 N/kg

Use the equation:

gravitational potential energy = mass × gravitational field strength × height

Figure 7 shows a boat pulling a person parasailing.

Figure 7 is not drawn to scale.

Write down the equation which links gravitational field strength (g), mass (m) and weight (W).

The weight of the person is 735 N.

Calculate the mass of the person.

gravitational field strength = 9.8 N/kg

The person is moving at a constant velocity.

What is the resultant force acting on the person?

The magnitude of the force of the rope on the person is 5000 N.

Draw a vector diagram on Figure 8 to determine the vertical component and the horizontal component of the force on the rope.

Figure 8

The force of the rope on the person parasailing is 5000 N.

The boat pulls the person for a horizontal distance of 1000 m.

The work done by the force of the rope on the person is not calculated by multiplying 5000 N by 1000 m.

Explain how the work done by the force of the rope on the person could be calculated.

Figure 5 shows a car travelling at a constant speed on a straight, level road.

Draw an arrow on Figure 5 to show the direction of the force of air resistance on the car.

The mass of the car is 850 kg

Calculate the weight of the car.

Use the equation:

weight = mass × gravitational field strength

gravitational field strength = 9.8 N/kg

What is the direction of the normal contact force of the road on the wheels?

Tick (✓) one box.

The car is travelling at constant speed.

The resultant force on the car is zero.

How does the size of the normal contact force of the road on the wheels compare with the weight of the car?

Tick (✓) one box.

A car is travelling at a constant speed.

A constant braking force of 5100 N is applied by the brakes.

The car decelerates and stops.

The braking distance is 38 m

Calculate the work done by the braking force.

Choose the unit from the box.

Use the equation: work done = force × distance

Which two factors affect braking distance?

Tick (✓) two boxes.

The distance a car travels during the driver’s reaction time is called the thinking distance.

Which factor affects thinking distance?

Tick (✓) one box.

Figure 6 shows a sketch graph of how thinking distance varies with speed.

Which term describes the relationship between thinking distance and speed?

Tick (✓) one box.

Figure 9 shows a car travelling at a constant speed on a straight, level road.

Figure 10 shows a free body diagram for the car.

Explain how the free body diagram shows that the resultant force on the car is zero.

Table 4 shows how the braking distance of a car varies with the speed of the car.

A car decelerates from 24 m/s to 0 m/s

The work done by the brakes in decelerating the car is 210 000 J

Determine the mean braking force.

Use information from Table 4.

Give your answer to 2 significant figures.

The braking distance is directly proportional to the kinetic energy of the car.

Explain how the data in Table 4 supports this relationship.