KE, GPE & EPE (AQA GCSE Combined Science: Trilogy: Physics): Revision Note

The diagram below shows a student before and after a bungee jump. The bungee cord has an unstretched length of 30.0 m.

The mass of the student is 60.0 kg. The gravitational field strength is 9.8 N/kg.

Calculate:

a) The change in gravitational potential energy of the student at 30.0 m

b) The maximum change in the gravitational potential energy of the student

c) The speed of the student after falling 30.0 m if 90% of the energy in the student's gravitational potential store is transferred to the student's kinetic store

d) The spring constant of the bungee cord if all the energy in the gravitational potential store of the student is transferred to the elastic potential store of the bungee cord

Answer:

Part (a)

Step 1: List the known quantities

• Mass of the student, m = 60.0 kg

• Gravitational field strength, g = 9.8 N/kg

• Change in height, h = 30.0 m

Step 2: Write down the equation for gravitational potential energy

$E_p=m\times g\times h$

Step 3: Substitute the known values and calculate the change in gravitational potential energy

$E_p=60.0\times 9.8\times 30.0$

$E_p=17 640\text{ J}$

Part (b)

Step 1: List the known quantities

• Mass of the student, m = 60.0 kg

• Gravitational field strength, g = 9.8 N/kg

• The maximum change in height is the distance from the bridge to the lowest point of the jump, h = 75.0 m

Step 2: Substitute the known values into the same equation and calculate the maximum change in gravitational potential energy

$E_p=m\times g\times h$

$E_p=60.0\times 9.8\times 75.0$

$E_p=44 100\text{ J}$

Part (c)

Step 1: List the known quantities

• Mass of the student, m = 60.0 kg

• Change in gravitational potential energy after falling 30.0 m, E_p = 17 640 J (from part (a))

Step 2: Calculate the energy transferred to the kinetic store

• Only 90% of the energy in the gravitational potential store is transferred to the kinetic store, so multiply by 0.9 (90% as a decimal):

$E_k=0.9\times 17 640$

$E_k=15 876\text{ J}$

Step 3: Write down the equation for kinetic energy

$E_k=\frac{1}{2}\times m\times v^2$

Step 4: Rearrange the equation to make speed the subject

• Multiply both sides by 2:

$m\times v^2=2\times E_k$

• Divide both sides by m:

$v^2=\frac{2\times E_k}{m}$

• Take the square root of both sides:

$v=\sqrt{\frac{2\times E_k}{m}}$

Step 5: Substitute the known values and calculate the speed

$v=\sqrt{\frac{2\times 15 876}{60.0}}$

$v=\sqrt{529.2}$

$v=23.0\text{ m/s}$

Part (d)

Step 1: List the known quantities

• Maximum change in gravitational potential energy, E_p = 44 100 J (from part (b))

• All of this energy is transferred to the elastic potential store of the cord, so at the lowest point E_e = 44 100 J

Step 2: Determine the extension of the bungee cord

• The extension is the stretched length minus the unstretched length of the cord:

$e=75.0-30.0$

$e=45.0\text{ m}$

Step 3: Write down the equation for elastic potential energy

$E_e=\frac{1}{2}\times k\times e^2$

Step 4: Rearrange the equation to make the spring constant, k, the subject

• Multiply both sides by 2:

$k\times e^2=2\times E_e$

• Divide both sides by e²:

$k=\frac{2\times E_e}{e^2}$

Step 5: Substitute the known values and calculate the spring constant

$k=\frac{2\times 44 100}{45.0^2}$

$k=\frac{88 200}{2025}$

$k=43.6\text{ N/m}$

A toy car is launched by a spring with k = 64 N/m compressed by 0.075 m. The car has mass 0.052 kg.

Calculate the maximum speed of the car.

Answer:

Step 1: List the known quantities

• Spring constant, k = 64 N/m

• Compression, e = 0.075 m

• Mass of the car, m = 0.052 kg

Step 2: Calculate the energy stored in the elastic potential store of the spring

• The elastic potential energy equation works for a compression in the same way as for an extension

$E_e=\frac{1}{2}\times k\times e^2$

$E_e=\frac{1}{2}\times 64\times 0.{075}^2$

$E_e=0.18\text{ J}$

Step 3: Link the elastic potential energy to the kinetic energy

• The car has its maximum speed when all the energy in the elastic potential store of the spring has been transferred to the kinetic store of the car:

$E_k=E_e=0.18\text{ J}$

Step 4: Rearrange the kinetic energy equation to make speed the subject

$E_k=\frac{1}{2}\times m\times v^2$

$v=\sqrt{\frac{2\times E_k}{m}}$

Step 5: Substitute the known values and calculate the speed

$v=\sqrt{\frac{2\times 0.18}{0.052}}$

$v=2.6\text{ m/s}$

Check how much of the energy is transferred before you copy a value across. If all the energy in the gravitational potential store is transferred to the kinetic store, the two values are equal. If only a fraction is transferred (for example 80%), multiply the energy transferred by that fraction. Use your answer from the previous part and show it, because marks can be awarded for following your own earlier answer through

Square the extension. Substitute the numbers first, then rearrange. Show every step: an answer with no working does not score full marks