Free AP Physics 1 Work, Energy, and Power

AP Physics 1 · Work, Energy, and Power · free lesson

Translational Kinetic Energy

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

The energy of motion

A rolling bowling ball can knock pins over, and a parked one cannot. A fast car does more damage in a crash than a slow one. Something about a moving object lets it do things, and that something grows with how fast and how massive the object is. Physics gives it a name and a formula.

Definition.

The translational kinetic energy of an object is the energy it has because its center of mass is moving. For an object of mass mm moving with speed vv,

K=12 mv2\begin{aligned} K &= \tfrac{1}{2}\,m v^2 \end{aligned}

Energy is measured in joules (J). One joule is one kilogram meter squared per second squared: 1 J=1 kg⋅m2/s21\ \mathrm{J} = 1\ \mathrm{kg\cdot m^2/s^2}.

A runner of mass 70 kg70\,\mathrm{kg} moving at 6.0 m/s6.0\,\mathrm{m/s} has a kinetic energy found by putting the numbers in the formula. Square the speed first, then multiply by the mass, then take half.

K=mv22=70 kg×(6 m/s)22=1260 J\begin{aligned} K &= \frac{mv^{2}}{2} \\ &= \frac{70\,\mathrm{kg} \times \left(6\,\mathrm{m/s}\right)^{2}}{2} \\ &= 1260\,\mathrm{J} \end{aligned}

The result is 1260 J1260\,\mathrm{J}. The unit comes out as joules by itself: kilograms times meters squared over seconds squared.

How kinetic energy depends on mass and speed

The formula has two inputs, and they do not matter in the same way. The kinetic energy is proportional to the mass: twice the mass, twice the energy. But it is proportional to the square of the speed: twice the speed, four times the energy; three times the speed, nine times the energy. Move the slider to change the speed of a 2.0 kg cart. The bar shows its kinetic energy, drawn to the same scale every time.

Going from 4 m/s to 8 m/s doubles the speed, and the bar becomes four times as long. This is why a car at highway speed is so much more dangerous than the same car at city speed.

The graph shows kinetic energy against speed for two objects, one of mass 2.0 kg2.0\,\mathrm{kg} and one of mass 4.0 kg4.0\,\mathrm{kg}. The blue curve is the lighter object and the red curve the heavier one. Both curves are parabolas starting at the origin: at any speed the heavier object's curve is twice as high, because its mass is twice as large.

Read the blue curve: at 5 m/s the 2.0 kg object has 25.0 J25.0\,\mathrm{J}, and at 10 m/s it has 100 J100\,\mathrm{J}. The speed doubled and the energy went up by a factor of four. At a fixed speed the graph of KK against mm is a straight line through the origin instead, because KK is proportional to mm.

Rule.

Scaling. If the mass is multiplied by a factor aa and the speed by a factor bb, the kinetic energy is multiplied by ab2a b^2.

K1=mv22=900 kg×(10 m/s)22=45000 JK2=m×(2v)22=900 kg×(2×10 m/s)22=180000 J\begin{aligned} K_1 &= \frac{mv^{2}}{2} \\ &= \frac{900\,\mathrm{kg} \times \left(10\,\mathrm{m/s}\right)^{2}}{2} \\ &= 45000\,\mathrm{J} \\ K_2 &= \frac{m \times (2v)^{2}}{2} \\ &= \frac{900\,\mathrm{kg} \times (2 \times 10\,\mathrm{m/s})^{2}}{2} \\ &= 180000\,\mathrm{J} \end{aligned}

A 900 kg900\,\mathrm{kg} car at 10 m/s10\,\mathrm{m/s} has 45000 J45000\,\mathrm{J}. At twice that speed it has 180000 J180000\,\mathrm{J}, four times as much.

Kinetic energy is a scalar

The formula has only the mass and the speed in it. A mass is a scalar, and the speed is the magnitude of the velocity, so the kinetic energy has a size and no direction: it is a scalar. It is never negative, since v2v^{2} is never negative. Two carts of the same mass moving at the same speed in opposite directions have different velocities and the same kinetic energy. Kinetic energies add as plain numbers, with no regard to direction.

Both 1.5 kg1.5\,\mathrm{kg} carts have KK = 12.0 J12.0\,\mathrm{J}. A cart at −4.0-4.0 m/s does not have a negative kinetic energy: the sign of a velocity component is a direction, and direction does not enter KK. The same holds in two dimensions: a ball with velocity components vxv_x and vyv_y has speed v=vx2+vy2v = \sqrt{v_x^2 + v_y^2} and kinetic energy 12mv2\tfrac12 m v^2, whatever way it is moving.

Kinetic energy depends on the observer

Speed is measured relative to a frame of reference, and different observers can measure different speeds for the same object, so they find different kinetic energies. Neither is wrong: kinetic energy is a quantity that depends on the frame.

A passenger of mass 60 kg60\,\mathrm{kg} walks toward the front of a train at 1.5 m/s1.5\,\mathrm{m/s} relative to the train. The train moves at 20 m/s20\,\mathrm{m/s} relative to the ground. A person on the platform sees the passenger move at 21.5 m/s21.5\,\mathrm{m/s}. A person sitting on the train sees 1.5 m/s1.5\,\mathrm{m/s}. Each uses the same formula with the speed that observer measures.

Kg=mvg22=60 kg×(21.5 m/s)22=13900 JKt=mvw22=60 kg×(1.5 m/s)22=67.5 J\begin{aligned} K_g &= \frac{mv_g^{2}}{2} \\ &= \frac{60\,\mathrm{kg} \times \left(21.5\,\mathrm{m/s}\right)^{2}}{2} \\ &= 13900\,\mathrm{J} \\ K_t &= \frac{mv_w^{2}}{2} \\ &= \frac{60\,\mathrm{kg} \times \left(1.5\,\mathrm{m/s}\right)^{2}}{2} \\ &= 67.5\,\mathrm{J} \end{aligned}

The ground observer finds 13900 J13900\,\mathrm{J} and the train observer finds 67.5 J67.5\,\mathrm{J}. An observer who rides along with the passenger at 21.5 m/s21.5\,\mathrm{m/s} sees the passenger at rest and finds a kinetic energy of zero. Which value is "the" kinetic energy? None of them alone: each is the kinetic energy relative to that observer's frame, and a problem should say which frame it uses. Unless a problem says otherwise, the frame is the ground.

Warning.

Three traps. Doubling the speed quadruples the kinetic energy; it does not double it. Kinetic energy is never negative and has no direction, so a velocity of −4-4 m/s does not give a negative KK. And the kinetic energy of one object is not the same for every observer: always ask which frame the speed was measured in.

Note.

This is the kinetic energy of the object's center of mass moving as a whole: translational kinetic energy. An object can have other energy too, from spinning or from vibrating inside; those are not part of 12mv2\tfrac12 m v^2.

Worked examples

Worked examples

Six examples of the kinds the AP exam asks on this topic: a calculation, a change in the speed and mass, finding a speed from an energy, reading a graph, a claim about direction, and a claim about observers.

Example.

Example 1. A calculation

A sprinter of mass 55 kg55\,\mathrm{kg} runs at a steady 9.0 m/s9.0\,\mathrm{m/s} along a straight track. What is the sprinter's translational kinetic energy relative to the track?

Solution

Use K=12mv2K = \tfrac12 m v^2, squaring the speed before multiplying.

K=mv22=55 kg×(9 m/s)22=2230 J\begin{aligned} K &= \frac{mv^{2}}{2} \\ &= \frac{55\,\mathrm{kg} \times \left(9\,\mathrm{m/s}\right)^{2}}{2} \\ &= 2230\,\mathrm{J} \end{aligned}
Result.

The sprinter's kinetic energy is 2230 J2230\,\mathrm{J}.

Example.

Example 2. What changes if the speed and mass change?

Car 1 has a mass of 1000 kg1000\,\mathrm{kg} and a speed of 8.0 m/s8.0\,\mathrm{m/s}. Car 2 has a mass of 2000 kg2000\,\mathrm{kg} and a speed of 4.0 m/s4.0\,\mathrm{m/s}. Which car has the greater kinetic energy, and in what ratio?

Solution

Car 2 is twice as massive but half as fast. Mass enters KK once and speed enters twice, so the speed should win. Calculate each.

K1=m1v122=1000 kg×(8 m/s)22=32000 JK2=m2v222=2000 kg×(4 m/s)22=16000 J\begin{aligned} K_1 &= \frac{m_1v_1^{2}}{2} \\ &= \frac{1000\,\mathrm{kg} \times \left(8\,\mathrm{m/s}\right)^{2}}{2} \\ &= 32000\,\mathrm{J} \\ K_2 &= \frac{m_2v_2^{2}}{2} \\ &= \frac{2000\,\mathrm{kg} \times \left(4\,\mathrm{m/s}\right)^{2}}{2} \\ &= 16000\,\mathrm{J} \end{aligned}
Result.

Car 1 has the greater kinetic energy, 32000 J32000\,\mathrm{J} against 16000 J16000\,\mathrm{J}: 2.002.00 times as much. Doubling the mass doubles KK, halving the speed divides it by four, so the net factor for car 2 is 2×14=122 \times \tfrac14 = \tfrac12.

Example.

Example 3. Finding the speed from the energy

A baseball of mass 0.145 kg0.145\,\mathrm{kg} has a kinetic energy of 60 J60\,\mathrm{J} relative to the ground. How fast is it moving?

Solution

Solve K=12mv2K = \tfrac12 m v^2 for the speed: v=2K/mv = \sqrt{2K/m}.

v=2Km=2×60 J0.145 kg=28.8 m/s\begin{aligned} v &= \sqrt{\frac{2K}{m}} \\ &= \sqrt{\frac{2 \times 60\,\mathrm{J}}{0.145\,\mathrm{kg}}} \\ &= 28.8\,\mathrm{m/s} \end{aligned}
Result.

The baseball moves at 28.8 m/s28.8\,\mathrm{m/s}. Check: 0.145 kg0.145\,\mathrm{kg} and 28.8 m/s28.8\,\mathrm{m/s} in the formula give back 60.0 J60.0\,\mathrm{J}.

Example.

Example 4. Reading a graph

The graph shows the kinetic energy of a cart against its speed. The marked point is at a speed of 6.0 m/s6.0\,\mathrm{m/s} and an energy of 54 J54\,\mathrm{J}. What is the cart's mass? What would its kinetic energy be at 12 m/s12\,\mathrm{m/s}?

Solution

The graph is a parabola through the origin, as K=12mv2K = \tfrac12 m v^2 says. One point is enough to find the mass: m=2K/v2m = 2K/v^2.

m=2Kpvp2=2×54 J(6 m/s)2=3.00 kgK2=m×(2vp)22=3 kg×(2×6 m/s)22=216 J\begin{aligned} m &= \frac{2K_p}{v_p^{2}} \\ &= \frac{2 \times 54\,\mathrm{J}}{\left(6\,\mathrm{m/s}\right)^{2}} \\ &= 3.00\,\mathrm{kg} \\ K_2 &= \frac{m \times (2v_p)^{2}}{2} \\ &= \frac{3\,\mathrm{kg} \times (2 \times 6\,\mathrm{m/s})^{2}}{2} \\ &= 216\,\mathrm{J} \end{aligned}
Result.

The mass is 3.00 kg3.00\,\mathrm{kg}. At twice the speed the kinetic energy is 216 J216\,\mathrm{J}, four times the value at the marked point. Doubling the speed does not double the energy: the curve is not a straight line.

Example.

Example 5. Justifying a claim

Two identical 0.50 kg0.50\,\mathrm{kg} carts roll on a straight track. Cart A has a velocity of 3.0 m/s3.0\,\mathrm{m/s} to the right, and cart B has a velocity of −3.0 m/s-3.0\,\mathrm{m/s}. A student says: "Cart B has a negative kinetic energy, so its kinetic energy is less than that of cart A." Is the student right?

Solution

The sign of a velocity component gives a direction. Kinetic energy depends on the square of the speed, and a square is never negative. Calculate both.

Ka=mva22=0.5 kg×(3 m/s)22=2.25 JKb=mvb22=0.5 kg×(−3 m/s)22=2.25 J\begin{aligned} K_a &= \frac{mv_a^{2}}{2} \\ &= \frac{0.5\,\mathrm{kg} \times \left(3\,\mathrm{m/s}\right)^{2}}{2} \\ &= 2.25\,\mathrm{J} \\ K_b &= \frac{mv_b^{2}}{2} \\ &= \frac{0.5\,\mathrm{kg} \times \left(-3\,\mathrm{m/s}\right)^{2}}{2} \\ &= 2.25\,\mathrm{J} \end{aligned}
Result.

The student is wrong. Both carts have the same kinetic energy, 2.25 J2.25\,\mathrm{J}: they have the same mass and the same speed, and kinetic energy is a scalar with no direction. Their velocities differ, but their kinetic energies do not.

Example.

Example 6. Two observers

A tennis player on a platform sees a ball of mass 0.060 kg0.060\,\mathrm{kg} moving at 25 m/s25\,\mathrm{m/s} east. A passenger on a train moving east at 20 m/s20\,\mathrm{m/s} watches the same ball. Find the ball's kinetic energy for each observer. Which one is correct?

Solution

The passenger sees the ball move at the difference of the speeds, since both go east: 5.0 m/s5.0\,\mathrm{m/s}.

Kp=mvb22=0.06 kg×(25 m/s)22=18.8 JKt=mvr22=0.06 kg×(5 m/s)22=0.750 J\begin{aligned} K_p &= \frac{mv_b^{2}}{2} \\ &= \frac{0.06\,\mathrm{kg} \times \left(25\,\mathrm{m/s}\right)^{2}}{2} \\ &= 18.8\,\mathrm{J} \\ K_t &= \frac{mv_r^{2}}{2} \\ &= \frac{0.06\,\mathrm{kg} \times \left(5\,\mathrm{m/s}\right)^{2}}{2} \\ &= 0.750\,\mathrm{J} \end{aligned}
Result.

The platform observer finds 18.8 J18.8\,\mathrm{J} and the passenger finds 0.750 J0.750\,\mathrm{J}. Neither is wrong: kinetic energy depends on the frame of reference in which the speed is measured. An observer moving east at 25 m/s25\,\mathrm{m/s} would find a kinetic energy of zero.

Practice

Answer AP-style questions on translational kinetic energy. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise translational kinetic energy with new numbers