Unit 2 · Topic 2.8 Beta

Spring Forces

An ideal spring exerts a force proportional to its stretch or compression from its relaxed length, F_s = −kΔx, always back toward the relaxed length; the spring constant k (N/m) is the slope of force against stretch, and a hanging mass at rest stretches it by mg/k.

Practice 1: Creating RepresentationsPractice 2: Mathematical Routines

Share to Google Classroom (opens in a new tab)

Question set for this topic

Part 1 · Hook

Why this matters

Kitchen scales, car suspensions, trampolines and pogo sticks all rely on the same simple rule: stretch or squash a spring twice as far and it pushes back twice as hard. That one proportionality, Hooke's law, comes back again and again later in the course, so it is worth getting exactly right now.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If y is directly proportional to x and x triples, y

  1. triples
  2. stays the same
  3. is divided by 3
  4. is multiplied by 9
Show the answer

Direct proportion: same factor of change.

  • Correct: triples:
  • stays the same:
  • is divided by 3:
  • is multiplied by 9:

2. A block hangs at rest from a cord. The cord's pull on it is

  1. equal to its weight
  2. larger than its weight
  3. smaller than its weight
  4. zero
Show the answer

At rest, the net force is zero.

  • Correct: equal to its weight:
  • larger than its weight:
  • smaller than its weight:
  • zero:

Part 4 · See it

See it first

The same vertical spring beside a ruler, three times: relaxed; with a block of mass m, stretched by Δx; and with a block of mass 2m, stretched by 2Δx. Beside it, a graph of force on the spring against stretch is a straight line through the origin, labeled |F_s| = k|Δx|, with slope k.
Double the hanging weight, double the stretch. The slope of force against stretch is the spring constant. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A spring is stretched or compressed by Δx from its relaxed lengthit exerts a force on whatever is attached to it, toward its relaxed length
  2. The spring is ideal (massless, within its limit)the force is proportional to the displacement: |F_s| = k|Δx|
  3. The spring constant k is largerthe spring is stiffer: more force for the same stretch, less stretch for the same force
  4. A mass hangs at rest on the springkΔx = mg, so measuring the stretch measures the weight (a spring scale)

Part 6 · Key ideas

Key ideas

  • Hooke's law: Fs = −kΔx. Δx is the displacement of the spring's end from its relaxed length; the minus sign means the force points back toward the relaxed length.
  • A stretched spring pulls; a compressed spring pushes. The spring constant k, in N/m, measures stiffness.
  • An ideal spring is massless and obeys Hooke's law, so it pulls equally on whatever is at each end.
  • A graph of |Fs| against |Δx| is a straight line through the origin with slope k.

Part 7 · Misconception

A common mistake

The wrong idea: Δx in Hooke's law is the length of the spring.

What actually happens: Δx is how far the spring is stretched or compressed from its relaxed length. A 30 cm spring stretched to 34 cm has Δx = 0.04 m.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Stretching a spring with hanging masses

A student hangs a light spring from a clamp and hangs slotted masses from its lower end. For each mass, once the mass hangs at rest, she measures how far the spring has stretched from its relaxed length with a meterstick. Use g = 9.8 m/s².

Hanging mass and stretch of the spring
Hanging mass (kg)Stretch (m)
0.0500.020
0.1000.040
0.1500.061
0.2000.080
0.2500.100

1. When the 0.150 kg mass hangs at rest, how does the spring's upward force on it compare with its weight?

  1. They are equal in size
  2. The spring force is larger, since the spring is stretched
  3. The spring force is smaller, since part of the weight stretches it
  4. The spring force is zero, since the mass is at rest
Show the answer

The mass hangs at rest, so the net force on it is zero: the spring pulls up exactly as hard as Earth pulls down. That is why the spring force equals mg in this experiment.

  • Correct: They are equal in size: Right: translational equilibrium.
  • The spring force is larger, since the spring is stretched: A stretched spring pulls, but at rest it pulls with exactly mg; a larger pull would accelerate the mass upward.
  • The spring force is smaller, since part of the weight stretches it: The whole weight is balanced by the spring's pull; nothing is left over.
  • The spring force is zero, since the mass is at rest: At rest the spring must still hold the mass up; otherwise the mass would fall.

2. A graph of the spring force (the weight of the hanging mass) against the stretch is close to a straight line through the origin. Use the first and last rows to find its slope, the spring constant, in N/m.

Type a number in N/m.

Show the answer

Spring force = mg: 0.050 × 9.8 = 0.49 N and 0.250 × 9.8 = 2.45 N. Slope k = ΔF ÷ Δx = (2.45 − 0.49) N ÷ (0.100 − 0.020) m = 1.96 N ÷ 0.080 m = 25 N/m (24.5).

  • Answer: 25 N/m

3. Using the pattern in the table, how far would the spring stretch with a 0.400 kg mass hanging at rest, in meters?

Type a number in m.

Show the answer

The stretch is proportional to the force: 0.400 kg is twice 0.200 kg, so the stretch is twice 0.080 m = 0.16 m. With k: Δx = mg ÷ k = 3.92 N ÷ 24.5 N/m = 0.16 m.

  • Answer: 0.16 m

Diagram

A cart held against a compressed spring

A 0.50 kg cart on a level, frictionless track is attached to an ideal spring of spring constant 80 N/m whose other end is fixed to a wall. A student pushes the cart toward the wall, compressing the spring by 0.15 m, and holds it at rest. Take the direction away from the wall as positive.

cartrelaxed position of the spring endcompression

4. What is the spring force on the cart while the student holds it, including its sign (positive away from the wall), in newtons?

Type a number in N.

Show the answer

Hooke's law: F_s = −kΔx. The spring is compressed, so its end is displaced toward the wall: Δx = −0.15 m. F_s = −(80 N/m)(−0.15 m) = +12 N, away from the wall. A compressed spring pushes outward.

  • Answer: 12 N

5. What force does the student exert on the cart while holding it at rest?

  1. 12 N toward the wall
  2. 12 N away from the wall
  3. 24 N toward the wall
  4. No force, since the cart is at rest
Show the answer

The cart is at rest, so the student's push balances the spring's 12 N push away from the wall: 12 N toward the wall.

  • Correct: 12 N toward the wall: Right: equal and opposite to the spring force, on the same cart.
  • 12 N away from the wall: Away from the wall is the spring's push; the student must push the other way.
  • 24 N toward the wall: Only the spring's 12 N needs balancing.
  • No force, since the cart is at rest: At rest means zero net force, but the spring still pushes, so the student must push back.

6. The student lets go. What is the size of the cart's acceleration just after release, in m/s²?

Type a number in m/s^2.

Show the answer

Just after release the only horizontal force is the spring's 12 N: a = F ÷ m = 12 N ÷ 0.50 kg = 24 m/s², away from the wall.

  • Answer: 24 m/s^2

Part 9 · Summary

Summary

An ideal spring exerts a force proportional to its stretch or compression from its relaxed length, F_s = −kΔx, always back toward the relaxed length; the spring constant k (N/m) is the slope of force against stretch, and a hanging mass at rest stretches it by mg/k.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections