2.8 Spring Forces

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. The spring scale at the fruit stand

Read the story as text

At a fruit stand, the seller hangs a 2.0 kg bag of apples on the hook of a hanging scale. The hook drops a little, bounces up and down a few times, and then settles 4.0 cm lower than before. The pointer reads 19.6 N.

Inside the scale there is nothing but a spring. Hang a bag twice as heavy and the hook drops twice as far. Take the scale into an elevator that is speeding up, and the pointer reads more than 19.6 N, even though the apples did not change.

How hard does a spring push or pull, and what does a spring scale really measure?

2. A box on a spring

3. A soft spring and a stiff spring

Quick check. One bag of apples stretches a spring 3 cm. Mei hangs three identical bags on it. How far does it stretch now?

4. Who pulls the top of the scale?

5. The scale in the elevator

6. Two springs: side by side or end to end

Quick check. A bag hangs from a spring scale in an elevator that is speeding up on its way up. What does the scale read?

7. Check yourself

Think of your answer first, then tap to see it.

a) You stretch a spring. Which way does it pull on your hand?

Show answer

Back toward its relaxed length, opposite the stretch. A squeezed spring pushes back out.

b) You stretch a spring three times as far. How does its force change?

Show answer

It becomes three times as big, because F = k x.

c) A spring with k = 200 N/m is stretched 0.15 m. How hard does it pull?

Show answer

F = k x = 200 N/m × 0.15 m = 30 N.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. A spring pulls with 6 N when it is stretched 2 cm. How hard does it pull when stretched 6 cm?

2. You and a friend pull on opposite ends of a spring scale, each with 40 N. Nothing moves. What does the scale read?

3. The same bag hangs, still, from spring A and then from spring B. A stretches 2 cm, B stretches 5 cm. Which spring has the bigger spring constant k?

The idea in plain words

Every spring has a natural length: the length when nothing stretches or squeezes it. Stretch it and it pulls back. Squeeze it and it pushes back. The spring always tries to return to its natural length. That is why the spring force is called a restoring force.

For an ideal spring (light, and not stretched too far), the force is proportional to how far you stretch or squeeze it. This is Hooke's law:

Fs = −k Δx     size: |Fs| = k |Δx|

Trap: putting the spring's total length into kΔx (a 10 cm spring pulled to 25 cm used as Δx = 25 cm). Instead: Δx = new length − natural length = 25 − 10 = 15 cm.
natural length (Δx = 0, no force) stretched: Δx > 0, F pulls back (left) squeezed: Δx < 0, F pushes back (right)

Springs in a scale

A hanging scale or a bathroom scale is a spring with a pointer. The pointer shows how much the spring is stretched or squeezed, and the dial turns that into a force. So a scale reads the spring force, not your weight. The two are equal only when you are at rest or moving at constant velocity (no acceleration). In an elevator speeding up, the spring must push harder than your weight, so the scale reads more.

For a mass m hanging at rest on a spring: k Δx = m g, so

Δx = m g / k

A spring pulls on both ends

A light spring pulls on whatever is at each end with the same size of force (Newton's third law plus a massless spring). If you and a friend each pull an end of a spring scale with 50 N, it reads 50 N, not 100 N. A wall holding one end would do the same job as your friend.

Two springs together

Trap: mixing up the rules, for example adding the k values of springs in series. Instead: side by side (parallel) the k values add and the pair is stiffer; end to end (series) the stretches add and the pair is softer. Two 490 N/m springs: 980 N/m parallel, 245 N/m series.

Worked with numbers: the fish scale

Read the steps as text
  1. Find k first. The seller tests the scale with a 0.50 kg mass, which stretches it 1.0 cm = 0.010 m. k = F/Δx = (0.50 × 9.8 N)/(0.010 m) = 4.9/0.010 = 490 N/m. At rest, the spring force equals the weight of the test mass. Always change cm to m first.
  2. Weight of the 2.0 kg fish: m g = 2.0 × 9.8 = 19.6 N. This is the force the spring must supply once the fish hangs still.
  3. Stretch: Δx = m g / k = 19.6 / 490 = 0.040 m = 4.0 cm. At rest, net force = 0, so k Δx = m g.
  4. In an elevator speeding up at 2.0 m/s² (up positive): Fs − m g = m a, so Fs = m(g + a) = 2.0 × (9.8 + 2.0) = 23.6 N. Stretch = 23.6/490 = 4.8 cm. The net force must point up to make the fish speed up, so the spring must pull harder than the weight. The scale reads 23.6 N although the weight is still 19.6 N.
  5. Two of these springs in series: 1/keq = 1/490 + 1/490 → keq = 245 N/m, and the fish (at rest) stretches the pair 19.6/245 = 8.0 cm. Side by side (parallel): keq = 980 N/m and the stretch is 2.0 cm. In series each spring stretches 4.0 cm, so together 8.0 cm. In parallel each spring carries half the load, 9.8 N, so each stretches 2.0 cm.

Try it: animations

Predict first, then press Play. Watch the free body diagram change as the spring stretches.

1. Pull a spring at a steady speed (Hooke's law)

A hand pulls the free end of a spring to the right at a constant speed. Right is positive. The graph shows the force the spring exerts on the hand.

Free body diagram of the hand

The spring pulls the hand back (left) with k Δx. The hand moves at constant velocity, so the arm must pull right just as hard: net force zero.

2. Hang a mass on a spring scale

The mass is attached to the unstretched spring and let go. Up is positive and y = 0 is the bottom of the unstretched spring. A little friction in the scale (damping) makes it settle.

Free body diagram of the mass

While it bounces, the spring force is sometimes less and sometimes more than m g, so the scale reading swings. Only when the mass is still does the scale read exactly m g.

3. Two springs: series or parallel

A hand lowers a mass slowly onto two springs until the springs hold it all by themselves. Set "Arrangement" to 0 for series (end to end) or 1 for parallel (side by side).

Free body diagram of the mass

While the hand still helps, hand + springs = m g. At the end the springs carry everything: keq Δx = m g.

Examples

Example 1: how hard does it pull? basic

A spring with k = 200 N/m is stretched 15 cm from its natural length. How large is the spring force, and which way does it point?

Show solution
Read the steps as text
  1. Change units: 15 cm = 0.15 m. k is in N/m, so Δx must be in m.
  2. |Fs| = k |Δx| = 200 × 0.15 = 30 N. Hooke's law.
  3. Direction: back toward the natural length, opposite to the stretch. The spring force is a restoring force.
Trap: plugging 15 cm in as 15, which gives 200 × 15 = 3000 N. Instead: change to metres first: 15 cm = 0.15 m, so F = 200 × 0.15 = 30 N.

Example 2: k from a graph medium

A student measures the force needed to stretch a spring: 1.0 N at 2.0 cm, 2.0 N at 4.0 cm, 3.0 N at 6.0 cm. (a) Find k. (b) How far will the spring stretch when a 0.30 kg mass hangs from it at rest?

Show solution
Read the steps as text
  1. Plot F (vertical) against Δx (horizontal). The points lie on a straight line through the origin. Straight line through the origin = an ideal spring that obeys Hooke's law.
  2. (a) Slope = ΔF/Δx = (3.0 − 1.0) N / (0.060 − 0.020) m = 2.0/0.040 = 50 N/m. The slope of F vs Δx is k. Use two points far apart on the line.
  3. (b) Δx = m g / k = (0.30 × 9.8)/50 = 2.94/50 = 0.059 m ≈ 5.9 cm. At rest, spring force = weight.

Example 3: series and parallel AP

Springs with k₁ = 300 N/m and k₂ = 600 N/m hold a 4.0 kg mass at rest. Find the total stretch when they are (a) in series and (b) in parallel. In (b), assume the bar stays level, so both springs stretch the same amount.

Show solution
Read the steps as text
  1. Weight: 4.0 × 9.8 = 39.2 N. At rest, the spring system supplies 39.2 N upward.
  2. (a) Series: 1/keq = 1/300 + 1/600 = 2/600 + 1/600 = 3/600, so keq = 200 N/m. Stretch = 39.2/200 = 0.196 m ≈ 0.20 m. Check: each spring carries 39.2 N. Spring 1 stretches 39.2/300 = 0.131 m, spring 2 stretches 39.2/600 = 0.065 m, total 0.196 m.
  3. (b) Parallel: keq = 300 + 600 = 900 N/m. Stretch = 39.2/900 = 0.044 m. Check: spring 1 pulls 300 × 0.04356 = 13.1 N, spring 2 pulls 600 × 0.04356 = 26.1 N; total 39.2 N (keep the unrounded stretch until the end).

Example 4: bathroom scale in an elevator AP

A 60 kg student stands on a bathroom scale in an elevator. The scale's spring has k = 2.0×10⁴ N/m. The scale reads 540 N. (a) Find the elevator's acceleration (up positive). (b) How much is the spring squeezed? (c) Can you tell which way the elevator is moving?

Show solution
Read the steps as text
  1. Forces on the student: spring (normal) force 540 N up, weight 60 × 9.8 = 588 N down. The scale reading is the spring force.
  2. (a) Fnet = 540 − 588 = −48 N, so a = −48/60 = −0.80 m/s² (0.80 m/s² downward). Newton's second law on the student.
  3. (b) Δx = F/k = 540 / (2.0×10⁴) = 0.027 m = 2.7 cm. Hooke's law; the scale was squeezed less than the 588/2.0×10⁴ = 2.9 cm it would be at rest.
  4. (c) No. A downward acceleration means either moving up and slowing down, or moving down and speeding up. Forces tell you the acceleration, not the velocity.

Practice (AP style)

1. A spring is stretched 3.0 cm from its natural length and pulls with 6.0 N. If it is stretched 9.0 cm instead, it pulls with

Show answer

18 N. F ∝ Δx: three times the stretch, three times the force. (k = 6.0/0.030 = 200 N/m; 200 × 0.090 = 18 N.) 54 N would be true if F grew with Δx², which it does not.

2. A student graphs the force applied to a spring (vertical axis) against the spring's stretch (horizontal axis). The spring constant k is

Show answer

The slope. F = k Δx has the form y = (slope) x. The area under F vs Δx is the work done (energy, Unit 3), not k. For an ideal spring the intercept is zero.

3. A 0.50 kg mass hangs at rest from a spring with k = 100 N/m. The spring is stretched by

Show answer

Δx = m g/k = (0.50 × 9.8)/100 = 4.9/100 = 0.049 m. 0.005 m forgets g (uses 0.50/100).

4. Two students pull on opposite ends of a light spring scale. Each pulls with 50 N and the scale is at rest. The scale reads

Show answer

50 N. A scale reads the tension in its spring. Hanging it from a hook on the wall and pulling with 50 N gives the same reading: the wall pulls back with 50 N, just like the second student.

5. A block is pushed against a spring attached to a wall on the left, squeezing the spring. The block is then held still. The force the spring exerts on the block is

Show answer

A squeezed spring pushes outward, back toward its natural length: on the block, to the right. The block is at rest because the hand pushes left with the same size force, not because the spring force is zero.

6. A spring with spring constant k is cut into two equal halves. The spring constant of each half is

Show answer

2k. The whole spring is the two halves in series. Under a force F, each half stretches half as much as the whole spring, so each half needs twice the force per metre. Check with the series rule: 1/k = 1/kh + 1/kh = 2/kh, so kh = 2k.

7. A mass stretches one spring by 6.0 cm. The same mass hangs at rest from two identical copies of that spring side by side (parallel). The stretch is

Show answer

Parallel: keq = 2k, so Δx = m g/(2k) = half of 6.0 cm = 3.0 cm. Each spring carries half the weight. 12 cm is the series answer.

8. A hanging spring scale holds a bag of apples in an elevator. The scale reads less than the weight of the apples. The elevator could be

Show answer

Spring force < m g means the net force is down, so the acceleration is down. Moving up and slowing down has a downward acceleration. Constant speed or rest: the reading equals the weight. Speeding up while moving up: the reading is more than the weight.

9. (Short free response: data and a graph) A student hangs masses on a spring and measures the stretch:

Mass (kg)0.100.200.300.40
Stretch (m)0.0210.0400.0610.080

(a) Which quantities should be plotted to get a straight line whose slope is k? (b) Find k. (c) A second student plots stretch on the vertical axis and force on the horizontal axis. What does that slope mean?

Show answer
  1. (a) Plot the spring force F = m g (vertical) against the stretch Δx (horizontal). Forces: 0.98, 1.96, 2.94, 3.92 N. At rest, spring force = weight, and F = k Δx is a line through the origin with slope k.
  2. (b) Slope from the best-fit line (use far-apart points): (3.92 − 0.98)/(0.080 − 0.021) = 2.94/0.059 ≈ 50 N/m. (A best-fit line through the origin gives about 49 N/m; anything from 48 to 50 N/m is fine.)
  3. (c) The slope is Δx/F = 1/k (about 0.020 m/N): how many metres the spring stretches per newton.

Scoring idea: 1 point for F (or m g) vs Δx, 1 point for slope = k, 1 point for a value 48-50 N/m with units, 1 point for 1/k in (c).

10. (Short free response: reasoning) A 1.0 kg bag hangs from a kitchen spring scale (k = 400 N/m) inside an elevator. (a) How far is the spring stretched when the elevator is at rest? (b) The elevator speeds up upward at 1.5 m/s². Find the new stretch and the scale reading. (c) Explain in words why the reading changes even though the bag's weight does not.

Show answer
  1. (a) Δx = m g/k = 9.8/400 = 0.0245 m ≈ 2.5 cm.
  2. (b) Up positive: Fs − m g = m a → Fs = 1.0 × (9.8 + 1.5) = 11.3 N. Δx = 11.3/400 = 0.028 m ≈ 2.8 cm.
  3. (c) To speed up upward, the bag needs a net upward force. Only two forces act: gravity (unchanged, 9.8 N) and the spring. So the spring must pull harder than 9.8 N, which means it stretches more, and the scale shows the spring force.

Scoring idea: 1 point for (a), 1 for the Newton's second law equation with correct signs, 1 for 11.3 N and 2.8 cm, 1 for an argument that names net force upward and the unchanged weight.

11. (Short free response: qualitative/quantitative translation) A 0.30 kg mass hangs at rest from a spring with k = 49 N/m. (a) Derive an expression for the stretch Δx and calculate it. (b) Without new numbers, predict how the stretch changes if (i) the mass is doubled, (ii) two of these springs hold the mass side by side, (iii) two of these springs hold it end to end. (c) Calculate the three stretches and check your predictions.

Show answer
  1. (a) At rest: kΔx = mg, so Δx = mg / k = 0.30 × 9.8 / 49 = 0.060 m = 6.0 cm.
  2. (b) (i) Doubles (Δx is proportional to m). (ii) Halves: side by side the pair is twice as stiff. (iii) Doubles: end to end each spring stretches the full 6.0 cm.
  3. (c) (i) 0.60 × 9.8 / 49 = 12 cm. (ii) keq = 98 N/m: 2.94 / 98 = 3.0 cm. (iii) keq = 24.5 N/m: 2.94 / 24.5 = 12 cm. All match.

Point guide (4 points):

  • 1 point: Δx = mg/k = 6.0 cm.
  • 1 point: correct predictions with reasons.
  • 1 point: correct keq for parallel and series.
  • 1 point: 12 cm, 3.0 cm, 12 cm.

12. (Short free response: experimental design and analysis) Design and analyze an experiment to test whether a spring obeys Hooke's law and to find its spring constant. You have the spring, a stand, a metre stick and a set of 0.050 kg slotted masses. (a) Describe your procedure, including what you measure and how you reduce error. (b) A student's data: 0.050 kg: 1.0 cm; 0.100 kg: 2.0 cm; 0.150 kg: 3.0 cm; 0.200 kg: 4.0 cm; 0.250 kg: 5.6 cm (stretch measured from the natural length). What should be plotted, and what is k? (c) What does the last data point suggest? (d) Name one source of error and how to reduce it.

Show answer
  1. (a) Hang the spring from the stand and record its natural length. Add masses one at a time; after each, wait until it is still and measure the new length at eye level. Stretch = new length − natural length. Remove the masses and check the spring goes back to its natural length.
  2. (b) Plot the spring force F = mg (vertical) against the stretch in metres (horizontal). The first four points give F = 0.49, 0.98, 1.47, 1.96 N at 0.010 to 0.040 m: a straight line through the origin, so Hooke's law holds there. Slope k = (1.96 − 0.49) / (0.040 − 0.010) = 49 N/m.
  3. (c) For 0.250 kg, F = 2.45 N, and Hooke's law predicts 2.45 / 49 = 0.050 m = 5.0 cm, but the spring stretched 5.6 cm. The point lies off the line: the spring has gone past its ideal (linear) range. Leave it out of the fit and do not load the spring further.
  4. (d) Example: reading the metre stick at an angle (parallax); read at eye level, or use a pointer fixed to the bottom of the spring. Or: the mass bouncing; wait until it is still.

Point guide (5 points):

  • 1 point: measures stretch from the natural length for several known masses.
  • 1 point: an error-reducing step (eye level, repeat, check return to natural length).
  • 1 point: F vs stretch in metres, k = 49 N/m from the linear part.
  • 1 point: last point is off the line, so beyond the ideal range.
  • 1 point: a real source of error with a matching fix.

13. (Short free response: mathematical routines) A 0.50 kg block hangs at rest from a spring with k = 140 N/m. (a) Calculate the stretch of the spring. (b) A hand pulls the block down until the total stretch is 0.060 m and holds it there. Calculate the force the hand exerts. (c) The hand lets go. Calculate the size and direction of the block's acceleration just after release.

Show answer
  1. (a) kΔx = mg, so Δx = 0.50 × 9.8 / 140 = 0.035 m (3.5 cm).
  2. (b) Spring force = 140 × 0.060 = 8.4 N up. Weight = 4.9 N down. Balance: Fhand = 8.4 − 4.9 = 3.5 N downward.
  3. (c) Fnet = 3.5 N up, so a = 3.5 / 0.50 = 7.0 m/s² upward.

Point guide (3 points):

  • 1 point: 0.035 m.
  • 1 point: 3.5 N downward, from 8.4 N − 4.9 N.
  • 1 point: 7.0 m/s² upward.

14. (Short free response: translation between representations) A graph shows spring force (vertical axis) against stretch (horizontal axis) for two springs. Both lines are straight and start at the origin. Line A passes through (0.10 m, 20 N); line B passes through (0.10 m, 5.0 N). (a) Find the spring constant of each spring from the graph. (b) A 1.0 kg mass hangs at rest from each spring in turn. Calculate each stretch. (c) Describe in words the free body diagram of the mass hanging from spring B. (d) Describe a graph of the stretch of spring A against the hanging mass, and give its slope.

Show answer
  1. (a) k = slope. A: 20 / 0.10 = 200 N/m. B: 5.0 / 0.10 = 50 N/m.
  2. (b) Weight = 1.0 × 9.8 = 9.8 N. A: 9.8 / 200 = 0.049 m. B: 9.8 / 50 = 0.196 m.
  3. (c) Two arrows of equal length: spring force 9.8 N up and weight 9.8 N down (equal because the mass is at rest).
  4. (d) A straight line through the origin: Δx = (g / k) m, so the slope is 9.8 / 200 = 0.049 m/kg.

Point guide (4 points):

  • 1 point: k from slope, 200 N/m and 50 N/m.
  • 1 point: 0.049 m and 0.196 m.
  • 1 point: two equal, opposite arrows of 9.8 N.
  • 1 point: straight line through origin, slope 0.049 m/kg.

Common mistakes

Using the total length instead of the stretch

The mistake: a 20 cm spring stretched to 25 cm, and using Δx = 0.25 m.

Why it is wrong: Hooke's law uses the change from the natural length: Δx = 0.25 − 0.20 = 0.05 m.

How to spot it: ask "what is Δx when the spring is relaxed?" It must be zero, and so must the force.

Forgetting the direction (the minus sign)

The mistake: drawing the spring force in the direction of the stretch.

Why it is wrong: the spring force always points back toward the natural length. Stretched springs pull; squeezed springs push.

How to spot it: imagine letting go. The object would move back toward the relaxed position; that is the force direction.

"A scale measures weight"

The mistake: assuming the scale reading always equals m g.

Why it is wrong: a scale reads its spring force. It equals m g only when the acceleration is zero.

How to spot it: if the problem has an elevator, a bounce or any acceleration, write Fs − m g = m a before using the reading.

Doubling the reading when both ends are pulled

The mistake: two people each pull with 50 N, so "the scale reads 100 N".

Why it is wrong: a light spring has the same tension throughout. One end must be held by something for the spring to stretch at all.

How to spot it: replace one person with a wall. The reading must be the same.

Mixing up series and parallel

The mistake: thinking that adding a second spring always makes the system stiffer.

Why it is wrong: end to end (series), each spring carries the full load and the stretches add, so the pair is softer. Side by side (parallel), the load is shared, so the pair is stiffer.

How to spot it: keq in series must be smaller than the smaller k; in parallel it must be bigger than the bigger k.

Centimetres in Hooke's law

The mistake: k = 500 N/m and Δx = 4 cm, giving F = 2000 N.

Why it is wrong: k is per metre. 4 cm = 0.04 m, so F = 20 N.

How to spot it: a kitchen spring giving thousands of newtons is a red flag. Convert to metres first.