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Mechanics

Conservation of energy and dissipation

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Overview

Recall questions on the principle of conservation of energy, why a falling body's speed does not depend on its mass, and why dissipated energy cannot be recovered.

What you need to know

Answer each one before you reveal it.

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Numbers

Why does a body falling freely through a height reach a speed given by the square root of 2gΔh?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why does a body falling freely from rest through a height Δh reach a speed given by v = √(2gΔh)?

    Model answer

    All the gravitational potential energy lost becomes kinetic energy, so mgΔh = ½mv²; the mass cancels and rearranging gives v = √(2gΔh).

  2. 2

    Why does the mass of a freely falling body not affect the speed it reaches?

    Model answer

    The mass appears on both sides of mgΔh = ½mv² and cancels, so the speed depends only on g and the height fallen.

  3. 3

    A ball is dropped from a height of 5.0 m. Using energy conservation and ignoring air resistance, what is its speed just before it lands?

    Model answer

    The speed is v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.

  4. 4

    What energy transfers occur as a pendulum bob swings from its highest point down to its lowest point and up again?

    Model answer

    Gravitational potential energy is transferred to kinetic energy on the way down and back to gravitational potential energy on the way up, with a small amount transferred to internal energy by air resistance on each swing.

  5. 5

    Why does a roller coaster car never quite reach the height of the hill it started from?

    Model answer

    Some of the initial gravitational potential energy is transferred to internal energy and sound by friction and air resistance, so less is available to lift the car.

  6. 6

    A car of mass 1200 kg travelling at 20 m s⁻¹ is brought to rest in 40 m. Using energy considerations, what is the average braking force?

    Model answer

    The kinetic energy is ½ × 1200 × 20² = 2.4 × 10⁵ J, and this equals FΔs, so F = 2.4 × 10⁵/40 = 6.0 × 10³ N.

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