Kinetic energy, the work-energy theorem and power

Last week you learned to calculate work. This week you will see what that work is good for: net work changes the speed of an object. This relationship is called the work-kinetic energy theorem, and it is Newton’s second law in the language of energy. The benefit: in problems where time and acceleration are not asked, you find the speed or distance in one line without dealing with $F = ma$ and the kinematic equations.

Golden rules:

1. $K = \tfrac12 mv^2$. Kinetic energy is a scalar and is never negative.
2. $W_{net} = \Delta K$. If the net work is positive the object speeds up, if negative it slows down, if zero its speed does not change.
3. Power is how fast the work is done: $P = W/\Delta t = Fv$.
4. Convert the speed to m/s before putting it into the formula. Since it is squared, the error grows.

Kinetic energy

The energy an object has because of its motion is called kinetic energy:

$K = \tfrac12 mv^2 \qquad [K] = \text{kg}\cdot\text{m}^2/\text{s}^2 = \text{J}$

Since it grows with the square of the speed, direction does not matter. The same object moving at $+5$ m/s and $-5$ m/s has the same kinetic energy.

Worked example 1. A 60.0 g bullet moves at 300 m/s.

$K = \tfrac12(0.0600\ \text{kg})(300\ \text{m/s})^2 = 2.70\times10^3\ \text{J}$

Someone who does not convert grams to kilograms finds a result 1000 times larger. That number is usually waiting among the options.

The work-kinetic energy theorem

For an object that travels a distance $d$ under a constant net force, $F_{net} = ma$ and $v_f^2 = v_i^2 + 2ad$. Combine the two:

$F_{net}\,d = ma\,d = \tfrac12 m(v_f^2 - v_i^2)$

The result is valid for variable forces too:

$\boxed{W_{net} = K_f - K_i = \Delta K}$

Here $W_{net}$ is the sum of the works of all the forces acting on the object: applied force, friction, gravity, spring, all with their signs.

Solution method

1. Draw the free-body diagram, list the forces.
2. Find the work of each force (W = F d cos theta; perpendicular ones are 0).
3. Add: W_net.
4. Write ½ m vf² - ½ m vi² = W_net, solve for the unknown.

Worked example 2. A 4.00 kg box slides on a horizontal floor at 3.00 m/s. A horizontal force of 25.0 N in the direction of motion is applied over 5.00 m. $\mu_k = 0.300$. What is the final speed of the box?

Work of the applied force:  W_F = 25.0 · 5.00 = 125 J
Friction: f = 0.300 · 4.00 · 9.80 = 11.76 N
          W_f = -11.76 · 5.00 = -58.8 J
N and mg: perpendicular to the motion → 0
W_net = 125 - 58.8 = 66.2 J
½ (4.00) vf² = ½ (4.00)(3.00)² + 66.2 = 18.0 + 66.2 = 84.2 J
vf² = 42.1  →  vf = 6.49 m/s

PITFALL · forgetting the initial kinetic energy. If the object does not start from rest, $K_i \neq 0$. Whoever writes "$\tfrac12 mv_f^2 = W_{net}$" finds 5.75 m/s in the example. The theorem states a change.

Braking distance

For a skidding car, the only force that does work is friction: $-\mu_k mg\,d = 0 - \tfrac12 mv^2$. From this:

$d = \frac{v^2}{2\mu_k g}$

The mass cancels. If the speed doubles, the distance quadruples. This is the physics counterpart of why speed limits matter in traffic.

Net work is zero, but the forces are working

When you lift a box at constant speed, the work you do is $+mgh$ and the work of gravity is $-mgh$. The net work is zero, and the kinetic energy does not change. This does not mean “no work was done”. Each force has its own work, and the question says which one it asks about: “work done by you”, “work done by gravity” or “net work”. In the options these three numbers sit side by side. Underline the name of the force in the question stem.

TIP · kinematics or energy? If time does not appear in the question and is not asked, the work-energy theorem is usually the shortest way. If time is asked, go back to kinematics ($v = v_0 + at$).

Power

A motor that does the same work in 2 seconds is more powerful than one that does it in 20 seconds. Power is the work done per unit time:

$P_{avg} = \frac{W}{\Delta t} \qquad P = \frac{dW}{dt} \qquad [P] = \text{J/s} = \text{W (watt)}$

Since with a constant force $dW = F\,dx\cos\theta$, the instantaneous power is:

$P = Fv\cos\theta = \vec F\cdot\vec v$

Practical units: $1\ \text{hp} = 746\ \text{W}$, $1\ \text{kWh} = (1000\ \text{W})(3600\ \text{s}) = 3.6\times10^6\ \text{J}$. The kWh is a unit of energy, not power.

Worked example 3. An 800 kg elevator goes up at a constant 3.00 m/s (no friction). What is the power of the motor?

Constant speed → a = 0 → cable force T = mg = 800 · 9.80 = 7840 N
P = T v = 7840 · 3.00 = 2.35 × 10⁴ W = 23.5 kW

PITFALL · power is not zero at constant speed. At constant speed the net work is zero and the kinetic energy does not change. But the motor continuously replaces the energy that gravity (or friction) takes. The power of the motor is calculated as $F_{motor}\cdot v$.

Relativistic kinetic energy (short note)

$\tfrac12 mv^2$ gives a wrong result for objects approaching the speed of light ($c = 3.00\times10^8$ m/s). In special relativity (relativistic kinetic energy):

$K = (\gamma - 1)mc^2 \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

$mc^2$ is the rest energy, and $\gamma mc^2$ is the total energy.

Worked example 4. For $v = 0.800c$, $\gamma = 1/\sqrt{1 - 0.640} = 1.667$. $K = 0.667\,mc^2$. The classical formula would give $\tfrac12 m(0.8c)^2 = 0.320\,mc^2$: less than half. At low speeds ($v \ll c$) the two formulas give the same result, and in everyday problems keep using $\tfrac12 mv^2$.

Mixed exam examples

Worked example 5 (power on a hill). A 1200 kg car goes up a $5.00^\circ$ hill at a constant 20.0 m/s. Air and rolling resistance total 500 N. What is the power the motor delivers to the wheels?

Constant speed → net force along the incline is 0
F_motor = m g sin5 + F_resistance = 1200·9.80·0.0872 + 500
        = 1025 + 500 = 1525 N
P = F_motor · v = 1525 · 20.0 = 3.05 × 10⁴ W = 30.5 kW
On a flat road: P = 500 · 20.0 = 10.0 kW

The hill tripled the power. Most of the power goes to doing work against gravity, that is, to increasing the car’s potential energy (week 12).

Worked example 6 (compare the two methods). An object moving at 12.0 m/s slows down at a constant $3.00\ \text{m/s}^2$. How far does it go before stopping?

Kinematics:   0 = 12.0² - 2·3.00·d   →  d = 24.0 m
Work-energy:  -m·3.00·d = 0 - ½ m 12.0²  →  d = 144/6.00 = 24.0 m

The two ways give the same answer, because the work-energy theorem is already a combination of $F = ma$ and $v^2 = v_0^2 + 2a\Delta x$. The real advantage of the energy method shows when the force is not constant (spring, variable force) or the path is curved. There the kinematic formulas do not work, but work-energy does.

Common mistakes