3.1 - Translational Kinetic Energy

Introduction to Energy

Energy describes the capacity to do work. It comes in many forms, and here the focus is kinetic energy, the energy of motion.

Translational Kinetic Energy

Translational kinetic energy is the energy an object has because it’s moving through space:

K = \frac{1}{2}mv^2

where: -

K

= kinetic energy (joules, J) -

m

= mass (kg) -

v

= speed (m/s)

Notice the formula uses speed (

v

), not velocity. Speed is just the magnitude of velocity.

The formula

K = \frac{1}{2}mv^2

appears on the AP equation sheet exactly like this. In two or three dimensions,

v

is the magnitude of the velocity vector, so

v^2 = v_x^2 + v_y^2 + v_z^2

. Example: a 0.40 kg ball has velocity

\vec{v} = (3.0\hat{i} + 4.0\hat{j})

m/s. Its speed is

\sqrt{3.0^2 + 4.0^2} = 5.0

m/s, so

K = \frac{1}{2}(0.40)(5.0)^2 = 5.0

J. That’s one number with no direction. There’s no “

x

-component of kinetic energy” to report.

Finding Speed from Position Using Calculus

In calculus-based physics you might get a position equation

x(t)

and need to find kinetic energy, which means finding speed first.

Step 1: Take the derivative of position to get velocity.

v(t) = \frac{dx}{dt}

Step 2: Speed is the magnitude of velocity. For one-dimensional motion where velocity can be positive or negative:

\text{speed} = |v(t)|

Often the context makes clear which direction is positive, and if the motion is in the positive direction you can use

v(t)

directly.

Step 3: Plug into the kinetic energy formula.

K(t) = \frac{1}{2}m[v(t)]^2

Properties of Kinetic Energy

1. Kinetic energy is always positive.

K \geq 0

Three reasons: the speed gets squared (so even a negative velocity becomes positive, since

v = -5

m/s gives

v^2 = 25

m²/s²), the mass is positive (no negative mass in classical physics), and the

\frac{1}{2}

is positive. So kinetic energy is always positive or zero (zero when

v = 0

2. Kinetic energy is a scalar. Unlike velocity, which is a vector, kinetic energy has magnitude only and no direction. For example, velocity might be

\vec{v} = -5\hat{i}

m/s (a vector with direction), speed is

v = 5

m/s (scalar), and kinetic energy is

K = \frac{1}{2}m(5)^2

J (scalar). So there’s no such thing as “negative kinetic energy” or “kinetic energy pointing left.”

3. It depends on the reference frame. Different observers measure different kinetic energies. Take a ball thrown at 10 m/s inside a moving train: an observer on the train measures

v = 10

m/s and gets

K = \frac{1}{2}m(10)^2

, while an observer on the ground measures

v = 10 + v_{train}

and gets a different

K

. This ties back to relative motion from Unit 1.

With numbers: a 0.50 kg ball is thrown toward the front of a train at 8.0 m/s relative to the train, and the train moves at 12 m/s relative to the ground. A passenger measures

K = \frac{1}{2}(0.50)(8.0)^2 = 16

J. Someone standing on the ground sees the ball moving at

8.0 + 12 = 20

m/s and measures

K = \frac{1}{2}(0.50)(20)^2 = 100

J. A bird flying right alongside the ball at 20 m/s would measure

K = 0

. All three are correct. Just pick one frame and stick with it for the whole problem.

Example Problem: Kinetic Energy from a Position Function

A 150 kg box slides across a rough surface with position given by:

x(t) = 0.6t^3 + 3t

where

x

is in meters and

t

in seconds.

Find the box’s speed at $t = 0$ .

Find the box’s kinetic energy as a function of time.

Part (a): Speed at t = 0. Start with the position function and take the derivative using the power rule

\frac{d}{dt}(t^n) = nt^{n-1}

v(t) = 0.6(3)t^{2} + 3(1)t^{0} = 1.8t^2 + 3

t = 0

v(0) = 1.8(0)^2 + 3 = 3 \text{ m/s}

The speed at

t = 0

is 3 m/s, which is the initial speed. The box is already moving when the clock starts.

Part (b): Kinetic energy as a function of time. Using

K = \frac{1}{2}mv^2

with

v(t) = 1.8t^2 + 3

K(t) = \frac{1}{2}(150)[1.8t^2 + 3]^2 = 75[1.8t^2 + 3]^2

So:

K(t) = 75(1.8t^2 + 3)^2 \text{ J}

Leave it in this form. Don’t expand

(1.8t^2 + 3)^2

unless you’re specifically asked, because the factored form is cleaner and less error-prone, and expanding invites mistakes like writing

(a + b)^2 = a^2 + b^2

(wrong) instead of

(a + b)^2 = a^2 + 2ab + b^2

(right).

Optional expansion (not recommended unless asked):

(1.8t^2 + 3)^2 = (1.8t^2)^2 + 2(1.8t^2)(3) + (3)^2 = 3.24t^4 + 10.8t^2 + 9

so:

K(t) = 75(3.24t^4 + 10.8t^2 + 9) = 243t^4 + 810t^2 + 675 \text{ J}

Still, the factored form

K(t) = 75(1.8t^2 + 3)^2

J is preferred.

Units of Energy

The SI unit of energy is the joule (J):

1 \text{ J} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2

which comes straight from the kinetic energy formula, since

K = \frac{1}{2}mv^2

has units of kg·(m/s)² = kg·m²/s² = J. You might also run into the kilowatt-hour (kWh, for electrical energy), the calorie (cal, in nutrition and chemistry), and the electron-volt (eV, in atomic and particle physics), and all of them convert to joules.

Checking Your Answer

Dimensional analysis. Always confirm the units. For

K(t) = 75(1.8t^2 + 3)^2

t

is in seconds,

1.8t^2

carries units of m/s (from the velocity function), so

(1.8t^2 + 3)

is in m/s,

(1.8t^2 + 3)^2

is in m²/s², and the 75 carries units of kg (from

\frac{1}{2}m = \frac{1}{2}(150)

). Multiplying gives kg·m²/s² = J, which checks out.

Initial condition check. At

t = 0

K(0) = 75(1.8(0)^2 + 3)^2 = 75(3)^2 = 675 \text{ J}

And directly from

K = \frac{1}{2}mv^2

K(0) = \frac{1}{2}(150)(3)^2 = 675 \text{ J}

They match.

Kinetic Energy and Direction

Two objects with the same mass and speed have the same kinetic energy no matter which way they’re going. For example, object A (5 kg, velocity

+10\hat{i}

m/s) and object B (5 kg, velocity

-10\hat{i}

m/s) both have:

K = \frac{1}{2}(5)(10)^2 = 250 \text{ J}

The negative sign in the velocity vanishes when you square it, which is exactly why kinetic energy is always positive and has no direction.

Comparing Kinetic Energies

Because

K

depends on

v^2

, speed matters much more than mass. Doubling the mass doubles

K

, but doubling the speed multiplies

K

by 4, and tripling the speed multiplies it by 9.

Example: a 1500 kg car at 10 m/s has

K = \frac{1}{2}(1500)(10)^2 = 7.5 \times 10^4

J. At 20 m/s it has

K = \frac{1}{2}(1500)(20)^2 = 3.0 \times 10^5

J, four times as much. With the same braking force, the brakes have to remove four times the energy, and since the brakes’ work is force times distance, the car needs four times the stopping distance.

Graphs: for a fixed mass, a graph of

K

vs.

v

is a parabola opening upward from the origin, and a graph of

K

vs.

v^2

is a straight line through the origin with slope

\frac{1}{2}m

. The straight-line version is the one to use with lab data, since a slope is easy to read off a best-fit line.

Common Mistakes

Using velocity instead of speed. The formula is $K = \frac{1}{2}mv^2$ where $v$ is speed. Because $v$ gets squared this usually doesn’t bite you, unless you mishandle vector notation.

Forgetting to square the velocity. It’s $\frac{1}{2}mv^2$ , not $\frac{1}{2}mv$ .

Dropping the $\frac{1}{2}$ . The formula has a $\frac{1}{2}$ , not just $mv^2$ .

Not taking the derivative first. Given $x(t)$ , you have to find $v(t) = \frac{dx}{dt}$ before computing kinetic energy.

Misapplying the power rule. Differentiating $t^n$ , bring the exponent down and drop the power by one: $\frac{d}{dt}(t^3) = 3t^2$ , not $t^2$ or $3t$ .

Expanding binomials wrong. $(a + b)^2 = a^2 + 2ab + b^2$ , not $a^2 + b^2$ . If you don’t need to expand, don’t.

Wrong units. Kinetic energy is in joules (kg·m²/s²). Check the final units.

Thinking kinetic energy can be negative. It’s always $\geq 0$ . A negative answer means an error.

Confusing speed and velocity. Speed is the magnitude of velocity (always positive), velocity can be negative. Since $v^2$ is positive either way it often doesn’t matter, but know the difference.

Not simplifying before substituting. Simplify $\frac{1}{2}m$ first (like $\frac{1}{2}(150) = 75$ ) before dealing with the velocity function.

Botching the $t = 0$ evaluation. Remember $0^n = 0$ for any positive $n$ , so only the constant terms survive.

Mixing reference frames. With multiple observers, be clear which frame you’re in. Kinetic energy differs between frames.

Forgetting the absolute value for speed. If velocity can be negative, use $K = \frac{1}{2}m|v|^2$ , though since $(-v)^2 = v^2$ this usually takes care of itself.

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