Translational Kinetic Energy

A $6000\ \text{kg}$ garbage truck rolling at $10\ \text{m/s}$ and a $1500\ \text{kg}$ car at $20\ \text{m/s}$ need the same amount of energy taken away by their brakes to stop. The truck has four times the mass, the car has twice the speed, and speed counts squared. The quantity that makes those two equal is translational kinetic energy, the energy an object has because its center of mass is moving.

Kinetic energy is one-half m v squared

"Translational" means the motion of the object as a whole from place to place. A wheel spinning in place on an axle has no translational kinetic energy, because its center of mass isn't moving; the energy of its spin is rotational kinetic energy, which Topic 6.1 covers. A ball flying through the air can have both. This topic is about the first kind only, and for any object you model as a single particle it is the only kind there is.

Where does $\frac{1}{2}mv^2$ come from? It is the work you must do to bring an object from rest to speed $v$. Push a mass $m$ along a line with a net force $F$. Newton's second law says $F = m\frac{dv}{dt}$, and the chain rule turns $\frac{dv}{dt}$ into $\frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$. So the work done over a small displacement $dx$ is

and adding these up from rest to the final speed gives

The force doesn't have to be constant and the details of the push don't matter: any net force that brings $m$ from rest to $v$ transfers exactly $\frac{1}{2}mv^2$ of energy to it. Topic 3.2 turns this calculation into the work-energy theorem.

Kinetic energy is proportional to mass and to the square of speed:

• double the mass at the same speed: $K$ doubles;
• double the speed at the same mass: $K$ is multiplied by $2^2 = 4$;
• triple the speed: $K$ is multiplied by $9$.

That is why braking distance grows so fast with speed. A car at $30\ \text{m/s}$ has $\left(\frac{30}{20}\right)^2 = 2.25$ times the kinetic energy it has at $20\ \text{m/s}$, and with the same braking force it needs $2.25$ times the distance to dissipate it.

A scalar built from a vector

Velocity has a direction; kinetic energy does not. The formula uses $v^2 = \vec{v}\cdot\vec{v} = v_x^2 + v_y^2 + v_z^2$, the square of the speed, and every component enters squared. Two consequences follow.

Direction never matters. A $0.50\ \text{kg}$ ball moving at $4.0\ \text{m/s}$ has $K = \frac{1}{2}(0.50)(4.0)^2 = 4.0\ \text{J}$ whether it moves up, down, east or at any angle. A ball that leaves your hand at $8\ \text{m/s}$ upward and later passes the same point at $8\ \text{m/s}$ downward has the same kinetic energy both times, even though its velocity reversed.

Kinetic energies add without canceling. Two $2.0\ \text{kg}$ carts moving toward each other at $3.0\ \text{m/s}$ each have velocities that cancel when you add them, but their kinetic energies are both positive: the system has $2 \times \frac{1}{2}(2.0)(3.0)^2 = 18\ \text{J}$. The kinetic energy of a system of objects is the sum of the kinetic energies of its parts, and since no term can be negative, the total is zero only if every part is at rest.

The component form tells you how to handle a projectile. Gravity changes $v_y$ and leaves $v_x$ alone, so at the top of a projectile's flight $v_y = 0$ but $v_x$ is still there: the kinetic energy at the peak is $\frac{1}{2}mv_x^2$, which is zero only for a launch straight up. Worked example 3 builds the full graph of $K$ against time.

Different observers measure different kinetic energies

Speed is measured relative to something, so kinetic energy is too. A passenger asleep in a seat on a train moving at $25\ \text{m/s}$ has zero kinetic energy in the train's frame and $\frac{1}{2}mv^2$ with $v = 25\ \text{m/s}$ in the ground's frame. Neither number is wrong. An observer's frame of reference is part of the description of any kinetic energy you state, and when a problem doesn't say, the frame is the ground (or whatever surface the motion is measured against).

The change is not a simple shift. If two frames move at velocity $\vec{u}$ relative to each other, velocities differ by $\vec{u}$ (Topic 1.4), but kinetic energies don't differ by any fixed amount, because the velocity is squared after the shift. Worked example 2 puts numbers on this for three observers of one hockey puck.

What every observer does agree on is the bookkeeping. In any frame that isn't accelerating, the change in an object's kinetic energy equals the net work done on it, measured in that same frame. You pick one frame at the start of a problem and use it for every speed in it; mixing a speed measured from a moving cart with one measured from the ground is how energy calculations go wrong.

Worked examples

Worked exampleKinetic energy of a projectile over time

Two identical $0.50\ \text{kg}$ balls are launched from level ground at $20\ \text{m/s}$. Ball 1 goes straight up. Ball 2 is launched at $30^\circ$ above the horizontal. Air resistance is negligible, and the gravitational field strength is $9.8\ \text{m/s}^2$.

(a) Derive an expression for Ball 2's kinetic energy $K$ as a function of time $t$ while it is in the air, in terms of $m$, $v_0$, $\theta$, $g$ and $t$.

(b) Calculate Ball 2's minimum kinetic energy and the time at which it occurs.

(c) On one set of axes, sketch $K$ versus $t$ for both balls from launch until each returns to the ground.

(d) A student says Ball 2 must have more kinetic energy than Ball 1 when it lands, because Ball 2 is moving sideways as well as down. Do you agree? Justify your answer.

Show solution

(a) Gravity acts only vertically, so the horizontal velocity stays $v_x = v_0\cos\theta$ and the vertical velocity is $v_y = v_0\sin\theta - gt$. Kinetic energy depends on the square of the speed, which is the sum of the squared components:

$$K(t) = \frac{1}{2}m\left[(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2\right].$$

(b) The first term never changes, and the second is smallest (zero) when $v_y = 0$, at $t = \frac{v_0\sin\theta}{g} = \frac{(20)(0.50)}{9.8} = 1.02\ \text{s}$. There the speed is $v_x = 20\cos 30^\circ = 17.3\ \text{m/s}$, so

$$K_{min} = \frac{1}{2}(0.50\ \text{kg})(17.3\ \text{m/s})^2 = 75\ \text{J}.$$

The launch kinetic energy is $\frac{1}{2}(0.50)(20)^2 = 100\ \text{J}$, so Ball 2 never has less than three quarters of it.

(c) Both curves start at $100\ \text{J}$. Ball 1's kinetic energy is $\frac{1}{2}m(v_0 - gt)^2$: a parabola opening upward that touches zero at the top of the flight, $t = 2.04\ \text{s}$, and returns to $100\ \text{J}$ when the ball lands at $t = 4.08\ \text{s}$. Ball 2's is the same parabola shape, because expanding $(v_0\sin\theta - gt)^2$ gives the same $t^2$ term, $\frac{1}{2}mg^2t^2$, that Ball 1's has; its vertex sits at $75\ \text{J}$ and $t = 1.02\ \text{s}$, and it returns to $100\ \text{J}$ at landing, $t = 2.04\ \text{s}$. It looks flatter on the graph only because it covers a smaller stretch of the parabola: $25\ \text{J}$ of drop instead of $100\ \text{J}$. A sketch earns credit for the features: same starting value, Ball 1 reaching zero, Ball 2's nonzero minimum at half of Ball 1's peak time, and each curve symmetric about its minimum.

(d) Disagree. When Ball 2 lands, its vertical velocity is $-v_0\sin\theta$, the reverse of its launch value, so its speed is again $v_0 = 20\ \text{m/s}$. Ball 1 also lands at $20\ \text{m/s}$. Kinetic energy depends only on mass and speed, not on direction, so both land with $100\ \text{J}$. Moving sideways doesn't add extra kinetic energy; the horizontal and vertical components together make up the same speed.

Where students lose points