AP Physics 1 · Lesson 2

Friction, springs and ramps

Know Know it

Friction acts along a surface. Kinetic friction acts while surfaces slide, with size fk=μkFNf_k = \mu_kF_N, against the sliding. Static friction acts while they do not slide, and it is only as large as needed to prevent sliding, up to a maximum of μsFN\mu_sF_N. Usually μs>μk\mu_s > \mu_k, which is why things are harder to start than to keep moving.

Friction depends on the normal force, not directly on the weight. The normal force equals the weight only for an object on a level surface with no other vertical forces: pulling up on a rope reduces it, and pressing down increases it.

An ideal spring exerts Fs=kxF_s = kx, where x is the stretch or compression from its relaxed length and k, in N/m, is its spring constant. The force points back toward the relaxed length.

Static and kinetic friction
FeatureStatic frictionKinetic friction
When it actsSurfaces not slidingSurfaces sliding
SizeAnything from 0 up to μsFN\mu_sF_NμkFN\mu_kF_N
DirectionOpposite the way the object would slideOpposite the sliding

Worked example Easy

A 4.5 N weight hung from a spring stretches it 6.0 cm from its relaxed length and hangs at rest. What is the spring constant?

  1. 0.27 N/m
  2. 7.5 N/m
  3. 75 N/mAnswer
  4. 750 N/m

How to solve it

  1. At rest, the spring pulls up with a force equal to the 4.5 N weight.
  2. Convert the stretch: 6.0 cm = 0.060 m.
  3. k=F/x=4.5/0.060=75k = F/x = 4.5/0.060 = 75 N/m.

Why each choice is right or wrong

  • A. Incorrect. 0.27 N/m multiplies the force by the stretch; Hooke’s law gives k as the force divided by the stretch.
  • B. Incorrect. 7.5 N/m reads the 6.0 cm stretch as 0.60 m; 6.0 cm is 0.060 m.
  • C. Correct. At rest the spring force balances the weight, so k=F/x=4.5/0.060=75k = F/x = 4.5/0.060 = 75 N/m.
  • D. Incorrect. 750 N/m reads the stretch as 0.0060 m, ten times too small; 6.0 cm is 0.060 m.

Trap Static friction is not always $\mu_sF_N$

μsFN\mu_sF_N is only the most static friction can give. A box at rest on a level floor with nobody pushing it has no friction on it at all. Push gently and static friction grows to match your push, until your push exceeds μsFN\mu_sF_N and the box starts to slide.

Explain Explain it

Task verbs: calculate, explain

On a ramp at angle θ, tilt your axes to run along and perpendicular to the surface. Gravity splits into mgsin⁡θmg\sin\theta down the slope and mgcos⁡θmg\cos\theta into the surface, so on a plain ramp FN=mgcos⁡θF_N = mg\cos\theta.

For an object sliding down a ramp with kinetic friction, Newton’s second law along the slope gives ma=mgsin⁡θ−μkmgcos⁡θma = mg\sin\theta - \mu_kmg\cos\theta. The mass cancels, so the acceleration depends only on the angle and the coefficient.

Worked example Medium

A 5.0 kg toboggan slides down a snowy slope that makes a 25° angle with the horizontal. The coefficient of kinetic friction is 0.20. What is the size of its acceleration?

  1. 2.2 m/s²
  2. 2.4 m/s²Answer
  3. 4.1 m/s²
  4. 5.9 m/s²

How to solve it

  1. Along the slope: the gravity component mgsin⁡25∘mg\sin 25^\circ acts down the slope and kinetic friction μkmgcos⁡25∘\mu_kmg\cos 25^\circ acts up it.
  2. Divide by m: a=g(sin⁡25∘−μkcos⁡25∘)a = g(\sin 25^\circ - \mu_k\cos 25^\circ).
  3. a=9.8(0.4226−0.1813)=2.4a = 9.8(0.4226 - 0.1813) = 2.4 m/s²; the 5.0 kg mass does not matter.

Why each choice is right or wrong

  • A. Incorrect. 2.2 m/s² uses FN=mgF_N = mg, as on level ground; on the slope FN=mgcos⁡25∘F_N = mg\cos 25^\circ, so friction is smaller.
  • B. Correct. a=g(sin⁡25∘−0.20cos⁡25∘)=9.8(0.423−0.181)≈2.4a = g(\sin 25^\circ - 0.20\cos 25^\circ) = 9.8(0.423 - 0.181) \approx 2.4 m/s² down the slope.
  • C. Incorrect. 4.1 m/s² is gsin⁡25∘g\sin 25^\circ, which leaves out friction.
  • D. Incorrect. 5.9 m/s² adds friction to the downhill pull, but kinetic friction points up the slope, against the sliding.

Explain Your turn

A 1.5 kg cart on a level, frictionless track is attached to a horizontal spring with a spring constant of 60 N/m. The cart is pulled 0.20 m from the spring’s relaxed length and released. Calculate the spring force on the cart and the cart’s acceleration just after release. Then explain how the cart’s acceleration changes as it moves back toward the relaxed position.

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Compare with a model answer

A full answer does this

  • Gives the spring force just after release as 12 N
  • Gives the acceleration just after release as 8.0 m/s²
  • Says the acceleration decreases as the cart moves toward the relaxed position
  • Gives the reason: the stretch gets smaller, so the spring force gets smaller

Model answer

Just after release the stretch is 0.20 m, so Fs=kx=(60)(0.20)=12F_s = kx = (60)(0.20) = 12 N, pointing back toward the relaxed position. The track is level and frictionless, so this is the net force, and a=F/m=12/1.5=8.0a = F/m = 12/1.5 = 8.0 m/s². As the cart moves back, the stretch x shrinks, so the spring force kxkx shrinks with it, and so does the acceleration kx/mkx/m. At the relaxed position the stretch is zero, so the force and the acceleration are both zero there.

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