AP Physics 1 · Lesson 2

Potential energy and conservation of energy

Know Know it

Potential energy belongs to a system of interacting objects, not to one object. Near Earth’s surface the object-Earth system has ΔUg=mgΔy\Delta U_g = mg\Delta y, and a spring stores Us=12kx2U_s = \tfrac{1}{2}kx^2. You may put the zero of UgU_g anywhere, because only changes matter.

If no outside forces do work on a system and no friction acts inside it, its mechanical energy K+UK + U stays constant: Ki+Ui=Kf+UfK_i + U_i = K_f + U_f. Energy only shifts between kinetic and potential forms.

If friction acts, mechanical energy is not conserved, but total energy still is: the lost mechanical energy shows up as thermal energy, ΔEth=fkd\Delta E_{\text{th}} = f_kd.

Where the energy is
EnergyFormulaBelongs to
KineticK=12mv2K = \tfrac{1}{2}mv^2Each moving object
Gravitational potentialΔUg=mgΔy\Delta U_g = mg\Delta yThe object-Earth system
Elastic potentialUs=12kx2U_s = \tfrac{1}{2}kx^2The spring
ThermalRises by fkdf_kd with sliding frictionThe surfaces that rub

Worked example Easy

The spring in a toy dart launcher has a spring constant of 400 N/m. How much elastic potential energy does it store when it is compressed 5.0 cm?

  1. 0.50 JAnswer
  2. 1.0 J
  3. 20 J
  4. 5,000 J

How to solve it

  1. Convert: 5.0 cm = 0.050 m.
  2. Us=12kx2=12(400)(0.0025)U_s = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(400)(0.0025).
  3. Us=0.50U_s = 0.50 J.

Why each choice is right or wrong

  • A. Correct. Us=12kx2=12(400)(0.050)2=0.50U_s = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(400)(0.050)^2 = 0.50 J.
  • B. Incorrect. 1.0 J leaves out the ½ in Us=12kx2U_s = \tfrac{1}{2}kx^2.
  • C. Incorrect. 20 J is kxkx, which is the spring force in newtons, not an energy.
  • D. Incorrect. 5,000 J uses the compression in centimeters, 5.0, instead of meters, 0.050.

Trap Mass cancels only when every term has it

For an object on a frictionless track, mghmgh and 12mv2\tfrac{1}{2}mv^2 both contain m, so the speed does not depend on the mass. Once a spring or a fixed amount of work supplies the energy, the mass no longer cancels: the same compressed spring launches a light block faster than a heavy one.

Explain Explain it

Task verbs: calculate, predict

Work every conservation problem the same way: choose the system, pick two moments, write the energy of each kind at each moment, and set the totals equal, adding any energy transferred in or out or turned into thermal energy.

To predict a speed or a height, you rarely need the path. A frictionless track can twist and turn; only the starting and ending heights and speeds matter.

Worked example Medium

A roller coaster car passes a point 25 m above the ground at 12 m/s. Ignoring friction and air resistance, how fast is it moving when it reaches a point 10 m above the ground?

  1. 17 m/s
  2. 21 m/sAnswer
  3. 25 m/s
  4. 29 m/s

How to solve it

  1. Car-Earth system, no friction: Ki+Ui=Kf+UfK_i + U_i = K_f + U_f.
  2. 12mvf2=12m(12)2+m(9.8)(25−10)\tfrac{1}{2}mv_f^2 = \tfrac{1}{2}m(12)^2 + m(9.8)(25 - 10), and m cancels.
  3. vf=144+294=438≈21v_f = \sqrt{144 + 294} = \sqrt{438} \approx 21 m/s.

Why each choice is right or wrong

  • A. Incorrect. 17 m/s ignores the 12 m/s the car already had and treats it as starting from rest.
  • B. Correct. 12v2=12(12)2+(9.8)(15)\tfrac{1}{2}v^2 = \tfrac{1}{2}(12)^2 + (9.8)(15), so v=144+294≈21v = \sqrt{144 + 294} \approx 21 m/s; the mass cancels.
  • C. Incorrect. 25 m/s uses the full 25 m height as the drop; the car descends only 25 - 10 = 15 m.
  • D. Incorrect. 29 m/s adds the speeds, 12 + 17; energies add, but speeds do not.

Predict Your turn

A skateboarder starts from rest at the top edge of a half-pipe whose two sides are each 3.2 m above its lowest point, and rolls down. Treat the half-pipe as frictionless and ignore air resistance. Predict her speed at the lowest point and how high she rises on the other side, and name the principle that justifies both predictions.

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Compare with a model answer

A full answer does this

  • Predicts a speed of 7.8 to 8.0 m/s at the lowest point
  • Predicts that she rises to 3.2 m on the other side
  • Names conservation of mechanical energy: kinetic plus potential energy stays constant

Model answer

With no friction or air resistance, the mechanical energy of the skateboarder-Earth system is conserved, so kinetic plus potential energy stays constant. At the lowest point all of the potential energy from the 3.2 m drop has become kinetic energy: 12v2=(9.8)(3.2)\tfrac{1}{2}v^2 = (9.8)(3.2), so v=62.7≈7.9v = \sqrt{62.7} \approx 7.9 m/s. On the far side she climbs until her kinetic energy is used up, which happens at the height she started from, 3.2 m.

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