Aim for FivePhysics C: Mechanics

Unit 3 · Topic 3.1

3.1 Translational Kinetic Energy

Kinetic energy, K=12mv2K = \frac{1}{2}mv^2, is the energy an object has because it's moving. It's a scalar that's never negative and grows with the square of speed, so doubling your speed quadruples it. Because speed depends on the reference frame, so does kinetic energy.

Key terms

  • kinetic energy
  • scalar
  • speed
  • reference frame
  • joule

What kinetic energy is

Translational kinetic energy is the energy of an object's motion from place to place:

K=12mv2K = \frac{1}{2}mv^2

Its unit is the joule (J), where 1 J = 1 kg·m²/s² = 1 N·m. A 1 kg ball moving at about 1.4 m/s has 1 J of kinetic energy.

"Translational" separates it from rotational kinetic energy, the energy of spinning, which comes in Unit 6. Here, objects are modeled as points, so all of their kinetic energy is translational.

A scalar that depends on speed squared

Kinetic energy is a scalar. It has no direction, so a ball moving east at 5 m/s and an identical ball moving west at 5 m/s have the same kinetic energy. Since mass is positive and v² can't be negative, K is never negative.

It depends on speed, not velocity: only the size of the velocity matters. An object moving in a circle at constant speed keeps a constant kinetic energy even though its velocity changes.

Because of the square, speed matters much more than mass. Double the mass and K doubles; double the speed and K quadruples; triple the speed and K grows ninefold. That's why stopping distance rises so steeply with speed: the brakes have to remove four times the energy at twice the speed.

For a system of several objects, the total kinetic energy is the sum of each object's kinetic energy. There's no canceling, unlike with momentum.

Kinetic energy depends on the reference frame

Speed is measured relative to an observer, so kinetic energy is too. A passenger sitting on a moving train has zero kinetic energy in the train's frame and a lot in the ground's frame. Neither value is wrong.

Within one problem, stick to one frame, usually the ground. Changes in kinetic energy are what matter for energy conservation, and every inertial observer agrees that energy is conserved, even though they disagree about the amounts.

Kinetic energy and momentum

Kinetic energy and momentum (Unit 4) both describe motion but aren't the same thing. Momentum, p⃗=mv⃗\vec{p} = m\vec{v}, is a vector and can cancel between objects; kinetic energy can't. They're linked by K=p22mK = \frac{p^2}{2m}, which is handy in collision problems. Two objects with the same momentum have different kinetic energies if their masses differ: the lighter one has more.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Doubling the speed

    A 1200 kg car speeds up from 10 m/s to 20 m/s. Find its kinetic energy before and after, and the change.

    Show the solution
    1. Step 1: Ki=12(1200)(10)2=60,000K_i = \frac{1}{2}(1200)(10)^2 = 60{,}000 J = 60 kJ.
    2. Step 2: Kf=12(1200)(20)2=240,000K_f = \frac{1}{2}(1200)(20)^2 = 240{,}000 J = 240 kJ.
    3. Step 3: ΔK=240−60=180\Delta K = 240 - 60 = 180 kJ.
    4. Step 4: The trap is assuming the second 10 m/s costs the same energy as the first. Getting from 0 to 10 m/s takes 60 kJ, but from 10 to 20 m/s takes 180 kJ, three times as much.

    Answer: 60 kJ, then 240 kJ; the change is 180 kJ (four times the energy at twice the speed).

  2. Example 2Calculator allowed

    Same ball, two frames

    A 2.0 kg ball is thrown forward at 5.0 m/s relative to a train that moves at 20 m/s relative to the ground, in the same direction. Find the ball's kinetic energy in the train's frame and in the ground's frame.

    Show the solution
    1. Step 1: Train frame: v = 5.0 m/s, so K=12(2.0)(5.0)2=25K = \frac{1}{2}(2.0)(5.0)^2 = 25 J.
    2. Step 2: Ground frame: v = 5.0 + 20 = 25 m/s, so K=12(2.0)(25)2=625K = \frac{1}{2}(2.0)(25)^2 = 625 J.
    3. Step 3: The difference, 600 J, is more than the 400 J the ball would have at 20 m/s alone. Because K depends on the square of the total speed, you can't find the ground-frame value by adding kinetic energies.

    Answer: 25 J in the train's frame; 625 J in the ground's frame.

Common mistakes

  • Giving kinetic energy a direction or a negative value. It's a scalar and never negative.
  • Assuming K is proportional to v. It's proportional to v², so a speed ratio gets squared.
  • Adding kinetic energies of objects moving in opposite directions as if they cancel. Kinetic energies always add.
  • Mixing reference frames within one problem.

On the exam

  • Ratio questions are common: if the speed triples and the mass halves, what happens to K? Multiply: 12×32=4.5\frac{1}{2} \times 3^2 = 4.5 times as much.
  • Graph questions may ask for K against v (a parabola) or K against v² (a straight line with slope m/2), a linearization that shows up on lab questions.

Connected topics

Videos

Check yourself

4 questions on 3.1 Translational Kinetic Energy. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the kinetic energy of a 1500 kg car moving at 20 m/s?

Question 2 of 4Calculator allowed

A cyclist doubles her speed. By what factor does her kinetic energy change?

Question 3 of 4Calculator allowed

Cart A has mass m and speed 2v. Cart B has mass 2m and speed v. How does cart A's kinetic energy compare with cart B's?

Question 4 of 4Calculator allowed

A 2.0 kg bag sits on the seat of a train moving at a constant 10 m/s relative to the ground. What is the bag's kinetic energy measured by a passenger and by a person standing on the platform?

0 of 4 answered