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Work, Energy and Power

Conservation of energy and energy exchanges

  • 3 things to know
  • Unit AS 1

Overview

Recall questions on the statement of the principle, the energy equation for a falling body and the derivation of v = √(2gΔh), how air resistance changes the exchange, why ramp shape does not matter, using light gates to investigate the exchange, and where the energy goes when a ball bounces low.

What you need to know

Answer each one before you reveal it.

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Numbers

A stone is dropped from rest from a height of 20 m. Calculate its speed just before it lands, ignoring air resistance.

Numbers

A pendulum bob is released from a position 0.25 m above its lowest point. Calculate its speed at the lowest point, ignoring resistive forces.

Key term

Describe how light gates can be used to investigate the exchange between gravitational potential energy and kinetic energy for a falling body.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the principle of conservation of energy.

    Model answer

    Energy cannot be created or destroyed, only transferred from one form to another, so the total energy of a closed system stays constant.

  2. 2

    Write down the energy equation for a body falling freely from rest, and state what each side represents.

    Model answer

    The equation is mgΔh = ½mv², where the left-hand side is the gravitational potential energy lost and the right-hand side is the kinetic energy gained.

  3. 3

    Show that the speed of a body falling freely from rest through a height Δh is v = √(2gΔh), and comment on the role of mass.

    Model answer

    Setting mgΔh = ½mv² and cancelling m gives v² = 2gΔh, so v = √(2gΔh); the mass cancels, so all bodies falling freely through the same height reach the same speed.

  4. 4

    A stone is dropped from rest from a height of 20 m. Calculate its speed just before it lands, ignoring air resistance.

    Model answer

    The speed is v = √(2 × 9.81 × 20) = 19.8 m s⁻¹, and ½ × 19.81² = 196.2 J kg⁻¹, which equals 9.81 × 20, confirming the result.

  5. 5

    A pendulum bob is released from a position 0.25 m above its lowest point. Calculate its speed at the lowest point, ignoring resistive forces.

    Model answer

    The speed is v = √(2 × 9.81 × 0.25) = 2.21 m s⁻¹, and ½ × 2.215² = 2.45 J kg⁻¹, which equals 9.81 × 0.25, confirming the result.

  6. 6

    How does air resistance change the energy exchange for a falling object?

    Model answer

    Some of the gravitational potential energy lost is transferred to the internal energy of the air and the object rather than to kinetic energy, so the object arrives with less kinetic energy and a lower speed than free fall would predict.

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