Energy, work and power

4.5 Energy, work and power

Work done by a force

$20 \times 5 \times \cos 30^\circ = 86.6 \text{ J}$

Kinetic and potential energy

Only the vertical height matters for potential energy, whatever the path. A particle moving 6 m along a slope at 30° rises $6\sin 30^\circ = 3$ m.

The work-energy principle

The change in a system's energy equals the work done by the external forces.

$\text{work by driving force} - \text{work against resistance} = \text{gain in KE} + \text{gain in PE}$

Conservation of energy

If only gravity does work, with no friction or driving force, then kinetic energy plus potential energy stays constant.

$mgh = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v^2 = 2 \times 10 \times 5 \quad\Rightarrow\quad v = 10 \text{ m s}^{-1}$

The mass cancels. The track's shape does not matter, since the normal reaction does no work.

With friction

PE lost: $2 \times 10 \times 3 = 60$ J. Work against friction: $4 \times 6 = 24$ J.

$60 - 24 = \tfrac{1}{2}(2)v^2 \quad\Rightarrow\quad v = 6 \text{ m s}^{-1}$

Power

Power is the rate of doing work. The unit is the watt, W, and 1 W is 1 J s⁻¹.

Cars on the level

The driving force at 20 m s⁻¹ is $\dfrac{24000}{20} = 1200$ N. Then

$1200 - 400 = 800a \quad\Rightarrow\quad a = 1 \text{ m s}^{-2}$

At the greatest speed, the acceleration is zero, so the driving force equals the resistance:

$\frac{24000}{v} = 400 \quad\Rightarrow\quad v = 60 \text{ m s}^{-1}$

Cars on hills

The component of the weight along the slope either helps or opposes the car.

The total opposing force is $400 + 800 \times 10 \times 0.1 = 1200$ N, so

$\frac{24000}{v} = 1200 \quad\Rightarrow\quad v = 20 \text{ m s}^{-1}$

Choosing a method