AP PHYSICS 1 AND 2 › UNIT 2, FORCE AND TRANSLATIONAL DYNAMICS › TOPIC 2.8
Spring Forces
The first force whose size is decided by where the object is rather than by who is pushing. Measured from the relaxed length, pointing always back toward it, and linear all the way, which is what makes it the most useful force in physics.
2.8.A Describe the force an ideal spring exerts on an object, measure it from the relaxed length, and say which way it points.
The whole topic on one sheet, the same one handed out in class. Click it to open the full-size version, which prints cleanly on a single page.
1. A force that depends on where the object is
Every force so far has been set by something outside the object: how hard somebody pushes, how heavy it is, how rough the floor is. A spring is different. The force it exerts is decided by how far it has been stretched or compressed, and by nothing else.
Fs = −k Δx
An ideal spring has negligible mass and obeys that relation exactly. The constant k is measured in newtons per meter and describes the spring, not the object hanging on it: a large k means a stiff spring that gives very little for a large force.
2. Where Δx is measured from
From the relaxed length. Not from the floor, not from the ceiling, not from where the block happens to be sitting, and not from the position it hangs at once you have loaded it. The relaxed length is the length the spring has when nothing is pulling on it, and that is the origin for everything in this topic.
Getting this wrong does not produce an obviously silly answer, which is exactly what makes it dangerous. Mark the relaxed position on the diagram before you write anything down.
Hang a mass on a spring and read the stretch. Every combination lands on the same straight line through the origin, and the slope of that line is the spring constant. Change the spring and the line tilts; change the mass and you slide along it.
3. The minus sign is the physics
The force always points back toward the relaxed position. Stretch the spring and it pulls inward; compress it and it pushes outward. That is what the minus sign in the formula says: the force and the displacement always point opposite ways.
A force with this property is called a restoring force, and it is why springs oscillate. Pull a block out and release it, and the force pushes it back, but it arrives at the relaxed position moving, overshoots, and gets pushed back the other way. Nothing else in this unit does that.
Move the block to either side of the relaxed position and watch the arrow. It never points away from home. The graph is a straight line through the origin with a negative slope, and both halves of it are the same law.
4. Measuring a spring constant
The standard experiment: hang a series of known masses, record the stretch each time, plot force against stretch, and take the slope. It works because the relation is linear and passes through the origin, so a graph with a nonzero intercept is telling you something is wrong, usually that the stretch was measured from the wrong place.
A hanging mass at rest is also the simplest equilibrium problem in the unit. Two forces, weight down and spring up, and they balance:
k Δx = mgso Δx = mg / k at the hanging position
5. Reading it wrong
Using the length instead of the change in length. A spring that is 0.30 m long and was 0.25 m relaxed has Δx = 0.05 m, not 0.30 m.
Thinking a stiffer spring pulls harder. Only at the same stretch. A stiff spring under a given hanging weight is stretched less and pulls with exactly the same force, because the weight decides the force and the spring decides the stretch.
Dropping the direction. The formula gives a size; the diagram has to show it pointing back toward the relaxed position. On a free-body diagram it is one straight arrow like any other.
Check yourself
1. A spring is 0.24 m long when relaxed. A block stretches it to 0.31 m and the spring pulls with 14 N. Find the spring constant.
Δx = 0.31 − 0.24 = 0.07 m, so k = 14 / 0.07 = 200 N/m. Dividing by 0.31 instead gives 45 N/m, which is the error this question exists to catch.
2. A 2.5 kg mass hangs at rest from a spring with k = 350 N/m. Find the stretch, and state the two forces on the mass and how they compare.
At rest means equilibrium, so the spring force equals the weight: kΔx = mg, giving Δx = 24.5/350 = 0.070 m. The two forces are the weight down from the Earth and the spring force up, equal in size and opposite in direction. They are not a third-law pair, which is the trap from Topic 2.3.
3. A block on a frictionless table is attached to a spring and pulled 0.12 m from the relaxed position, then released. Describe the force and the acceleration at the moment of release, as the block passes the relaxed position, and 0.12 m on the other side.
At release the force is at its largest and points back toward the relaxed position, so the acceleration is largest there and the speed is zero. Passing the relaxed position the force is zero and so is the acceleration, and the speed is greatest. At 0.12 m on the far side the force is the same size as at the start but points the other way, and the block is momentarily at rest again. Largest force where the speed is zero is the pattern worth remembering.
4. Two students hang the same 1 kg mass on two different springs. One stretches 4 cm, the other 10 cm. Compare the spring constants and compare the forces the springs exert.
The forces are identical, 9.8 N each, because both are holding up the same weight in equilibrium. The constants are not: 9.8/0.04 = 245 N/m and 9.8/0.10 = 98 N/m. The stiffer spring is the one that stretched less, and stiffness has nothing to do with how hard it is pulling here.
5. A student plots force against the total length of a spring rather than against the change in length, and gets a straight line that does not pass through the origin. Explain the graph and say how to get the spring constant from it anyway.
The graph is F = k(L − L₀), a straight line with the same slope as before but shifted, crossing the horizontal axis at the relaxed length rather than at zero. The slope is still the spring constant, so it is recoverable, and the intercept is a bonus: it measures L₀. A nonzero intercept on a force against stretch graph, by contrast, means an error somewhere in the measurement.
Topic 2.9, Circular Motion. Every force in this unit is now available. The last topic takes an object that is going at a steady speed and still accelerating, and asks which of these forces is doing it.