AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/3/3-1)
Unit 3 · Topic 3.1
3.1 Translational Kinetic Energy
Kinetic energy is the energy an object has because it's moving: K = ½mv². It's a scalar measured in joules, it's never negative, and because speed is squared, small changes in speed make big changes in kinetic energy. Its value also depends on the reference frame you measure from.
Key terms
- kinetic energy
- scalar
- joule (J)
- reference frame
The formula
Translational kinetic energy is K = ½mv², where m is the mass in kilograms and v is the speed in m/s. The unit is the joule: 1 J = 1 kg·m²/s² = 1 N·m.
"Translational" means the energy of the whole object moving from place to place. Spinning objects also have rotational kinetic energy, which comes in Unit 6.
A scalar that's never negative
Kinetic energy has no direction. A 2 kg ball moving left at 3 m/s has the same kinetic energy (9 J) as one moving right or up at 3 m/s. Since v² and m are both positive, K is never negative. It's zero only when the object is at rest.
This is a key difference from momentum (4.1), which is a vector. Two carts moving toward each other can have a total momentum of zero, but their total kinetic energy is the sum of two positive numbers.
How K scales
Because speed is squared, kinetic energy is very sensitive to speed. Mass enters only to the first power.
| Change | Effect on K |
|---|---|
| double the mass | K doubles (×2) |
| double the speed | K quadruples (×4) |
| triple the speed | K is 9 times as large |
| halve the speed | K drops to ¼ |
| double both mass and speed | K is 8 times as large |
It depends on the observer
Speed depends on the reference frame (1.4), so kinetic energy does too. A suitcase on a moving train has zero kinetic energy relative to the train, but plenty relative to the ground. Different observers can measure different values, and each is correct in its own frame. Within one problem, measure every speed from the same frame.
A single object treated as a point (the object model) can only have kinetic energy. It can't store potential energy by itself, because potential energy belongs to a system of interacting objects (3.3). That's why choosing the system matters so much in this unit.
Kinetic energy on graphs
A graph of K against v is a parabola opening upward, starting at the origin. For an object speeding up from rest with constant acceleration, v grows in proportion to t, so K against t is also a parabola. But v² grows in proportion to distance traveled (from v² = 2ad), so K against position is a straight line. That straight line is a preview of the work–energy theorem in 3.2: a constant net force adds the same energy for every meter.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Kinetic energy of a car
A 1200 kg car travels at 25 m/s. Find its kinetic energy. At what speed would it have half as much kinetic energy?
Show the solutionHide the solution
- Step 1: K = ½mv² = ½(1200)(25)² = ½(1200)(625) = 375,000 J = 3.75 × 10⁵ J.
- Step 2: For half the kinetic energy, v² must be half as large, so v = 25 ÷ √2 ≈ 17.7 m/s.
- Step 3: Note that half the energy doesn't mean half the speed. Half the speed would leave only a quarter of the energy.
Answer: 3.75 × 10⁵ J; about 17.7 m/s
- Example 2Calculator allowed
Speeding up from 10 to 20 vs. from 20 to 30 (classic trap)
A 1.0 kg cart speeds up from 10 m/s to 20 m/s, and later from 20 m/s to 30 m/s. Both are 10 m/s gains. Which takes more energy, and by what factor?
Show the solutionHide the solution
- Step 1: First gain: ΔK = ½(1.0)(20² − 10²) = ½(400 − 100) = 150 J.
- Step 2: Second gain: ΔK = ½(1.0)(30² − 20²) = ½(900 − 400) = 250 J.
- Step 3: 250 ÷ 150 ≈ 1.67. The same increase in speed costs more energy at higher speed, because K depends on v², not v.
- Step 4: The trap is computing ½m(Δv)², which gives 50 J for both. Always find each K first and then subtract.
Answer: The 20 → 30 m/s gain needs 5/3 as much energy (250 J vs. 150 J)
- Example 3Calculator allowed
Same ball, two frames
A train moves east at 20 m/s. A passenger throws a 0.15 kg ball east at 10 m/s relative to the train. Find the ball's kinetic energy measured (a) by the passenger and (b) by someone standing on the ground.
Show the solutionHide the solution
- Step 1: (a) Relative to the train, v = 10 m/s: K = ½(0.15)(10)² = 7.5 J.
- Step 2: (b) Relative to the ground, v = 10 + 20 = 30 m/s (1.4): K = ½(0.15)(30)² = 67.5 J.
- Step 3: Both are correct. Kinetic energy depends on the frame of reference.
Answer: (a) 7.5 J; (b) 67.5 J
Common mistakes
- Giving kinetic energy a direction or a negative sign. It's a scalar and is never negative.
- Computing ½m(Δv)² for a change in kinetic energy. Use ½mv_f² − ½mv_i².
- Assuming that doubling the speed doubles K. It quadruples.
- Mixing reference frames within one problem, such as one speed relative to a train and another relative to the ground.
On the exam
- Ratio questions are common: if the speed triples, by what factor does the kinetic energy change? Square the speed factor and multiply by the mass factor.
- In free-response answers, name the system before you name its energies. A system of one object can only have kinetic energy; an object + Earth system can also have gravitational potential energy.
Connected topics
Videos
Check yourself
4 questions on 3.1 Translational Kinetic Energy. Pick an answer to see if you got it, and why.
What is the kinetic energy of a 1000 kg car moving at 20 m/s?
A 2.0 kg ball has 36 J of kinetic energy. What is its speed?
Cart X has twice the mass of cart Y but half its speed. How does cart X's kinetic energy compare with cart Y's?
A suitcase sits on the rack of a train moving at constant velocity. A passenger on the train and a person standing on the platform each describe the suitcase's kinetic energy. Which statement is correct?
0 of 4 answered