Work, Energy and Power

≈ 30 min

Work done by a constant force $\vec{F}$ through a displacement $\vec{d}$ is

W = Fd\cos\theta

where $\theta$ is the angle between force and displacement. Work is a scalar, measured in joules, and it is zero whenever the force is perpendicular to the motion.

Kinetic energy is $K = \tfrac{1}{2}mv^2$ , and the work–energy theorem states

W_{\text{net}} = \Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2

For a conservative force such as gravity we may define a potential energy, $U = mgh$ near the Earth's surface, and then mechanical energy $E = K + U$ is conserved when no non-conservative force (friction, air drag) does work.

Power is the rate of energy transfer:

P = \frac{W}{\Delta t} = Fv \quad (\text{for } \vec F \parallel \vec v)

Worked example. A $50\ \text{kg}$ student runs up a $4.0\ \text{m}$ staircase in $8.0\ \text{s}$ .

Work done against gravity:

W = mgh = (50)(9.8)(4.0) = 1960\ \text{J}

Average useful power:

P = \frac{W}{\Delta t} = \frac{1960}{8.0} = 245\ \text{W}

That is roughly the output of a small ceiling fan — a useful reminder of how modest sustained human power really is.

A shopper carries a $10\ \text{kg}$ box horizontally at constant speed for $20\ \text{m}$ . How much work does her upward carrying force do on the box?

$0\ \text{J}$

$200\ \text{J}$

$1960\ \text{J}$

$98\ \text{J}$

How much work is done in lifting a $20\ \text{kg}$ bag of maize meal a vertical height of $1.5\ \text{m}$ at constant speed? Take $g = 9.8\ \text{m/s}^2$ and give the answer in joules.

A $1000\ \text{kg}$ car speeds up from $10\ \text{m/s}$ to $20\ \text{m/s}$ on a level road. What is the net work done on the car, in joules?

A borehole pump raises $500\ \text{kg}$ of water through a vertical height of $12\ \text{m}$ in $60\ \text{s}$ . Taking $g = 9.8\ \text{m/s}^2$ and ignoring losses, what is the minimum power output of the pump, in watts?