Model elastic forces with F = kΔx, measure a spring constant and analyze a mass hanging from a spring

Learning goal: Describe how linearly elastic objects stretch or compress in proportion to the load; use F_elastic = kΔx with Δx measured from the relaxed position and the force directed toward that position; determine a spring constant from force-stretch data; draw free-body diagrams for a mass hanging from a spring at rest and, qualitatively, at points of its up-and-down motion; and calculate the force on a mass hanging in equilibrium.

What an elastic force is. Stretch a spring and it pulls back; compress it and it pushes back. The force an elastic object exerts always points toward its relaxed position, the length it has when nothing loads it. For many springs, rubber cords and thin rods, the size of that force is proportional to how far the object has been displaced from its relaxed position, as long as the displacement is not too large. This is Hooke's law:

F_elastic = kΔx,

where Δx is the displacement from the relaxed position (a stretch or a compression, in meters) and k is the spring constant in newtons per meter. The constant depends on the stiffness of the object: a stiff spring has a large k, a soft one a small k, and for a given spring k is the same for compression as for extension. A spring with k = 20 N/m needs 20 N to stretch it 1.0 m, or 2.0 N to stretch it 0.10 m. Two conditions matter. The relation holds only within the spring's linear range, up to its limit of proportionality; beyond that the force is no longer proportional to the displacement. That is a different thing from the elastic limit, the larger stretch beyond which the object no longer returns to its relaxed length and is permanently deformed: between the two limits a spring can respond non-linearly and still recover fully. And Δx is measured from the relaxed length, never from some other convenient mark.

Reading a force-stretch graph. Hang masses from a spring and measure the stretch. Suppose 0.100 kg stretches it 0.049 m, 0.200 kg stretches it 0.098 m, 0.300 kg 0.147 m and 0.400 kg 0.196 m. The force on the spring is the weight hanging from it: 0.98, 1.96, 2.94 and 3.92 N. Plot force against stretch: the points lie on a straight line through the origin, and the slope is k = 3.92 N ÷ 0.196 m = 20 N/m. A straight line through the origin is the graphical statement of proportionality. If the points curved away from the line at the top, the spring would be past its limit of proportionality, and only unloading it would show whether it had also passed its elastic limit; if the line did not pass through the origin, the spring might have coils pressed together that need a small force before any stretch begins.

A mass hanging in equilibrium. A 0.250 kg mass hangs at rest from the 20 N/m spring. Free-body diagram of the mass: weight 0.250 × 9.8 = 2.45 N downward; elastic force from the spring upward. At rest the net force is zero, so the elastic force equals 2.45 N, and the stretch is Δx = F/k = 2.45 ÷ 20 = 0.1225 m, about 12 cm. This position, where the spring's upward pull just balances the weight, is the equilibrium position. Reverse the calculation to weigh something: a spring scale with k = 250 N/m that stretches 0.040 m is exerting 250 × 0.040 = 10 N, so the object's mass is 10 ÷ 9.8 = 1.02 kg.

A fully worked model: forces at an instant during the motion. A 0.50 kg mass hangs from a spring with k = 35 N/m. The equilibrium stretch is 4.9 ÷ 35 = 0.14 m. Now the mass is pulled down until the stretch is 0.20 m and released. At that instant the spring's upward force is 35 × 0.20 = 7.0 N and the weight is 4.9 N downward, so the net force is 2.1 N upward and the acceleration is 2.1 ÷ 0.50 = 4.2 m/s² upward. The mass rises, and as it passes the equilibrium position the spring force is exactly 4.9 N, the net force is zero and the velocity is greatest. It keeps rising: above equilibrium the spring force is less than the weight, the net force is downward, and the mass slows, stops and comes back. The motion repeats. Only the net force at an instant is computed here; the motion itself is described qualitatively in this course.

Diagrams through one cycle. Consider a mass attached to an unstretched vertical spring, held so the spring is at its relaxed length, and released. Initial position: elastic force zero, only the weight acts, net force down, the mass starts to move downward. Halfway to equilibrium: spring force half the weight, net force still downward but smaller, the mass still speeds up. Equilibrium: spring force equals weight, net force zero, speed greatest. Halfway from equilibrium to the lowest point: spring force greater than weight, net force upward, the mass slows. Lowest point: velocity zero, spring force greatest, net force upward and largest, so the mass immediately starts back up. At every point the net force points toward equilibrium, which is why the motion oscillates about that position.

How far does it drop? The lowest point is not the equilibrium position; the mass overshoots. Released from the relaxed length, it falls to a stretch of 2mg/k, twice the equilibrium stretch, before turning back. One way to see this uses work from physical science: over the drop the weight does work mg × Δx, while the spring's force grows from 0 to kΔx and so has an average of ½kΔx, doing work −½kΔx²; the mass stops when these cancel, so mgΔx = ½kΔx² and Δx = 2mg/k. You do not need to derive it here, but you can test it, and it matters for design: a cord that would hold a load at rest with a stretch of 0.4 m will stretch about 0.8 m if the load is released at rest from the cord's relaxed length. A load that has already fallen before the cord goes taut stretches it further still, as the design below shows.

Design: a bungee for a flour bag. A 1.0 kg bag is to be dropped from a 3.0 m platform on a cord and must stop within 0.30 m of the floor, never touching it. Choose a cord of unstretched length 2.0 m. The bag then falls 2.0 m freely, and the cord must stretch no more than 0.8 m for the lowest point to be at 2.8 m, 0.2 m above the floor. The shortcut 2mg/k does not apply here, because the bag reaches the cord already moving. Use the same work argument over the whole 2.8 m drop instead: the weight does work mg(2.0 + 0.8), the cord does work −½k(0.8)², and the bag is at rest at the top and the bottom, so 9.8 × 2.8 = ½k × 0.64 and k = 2 × 9.8 × 2.8 ÷ 0.64 = 86 N/m. At the lowest point the elastic force is 85.75 × 0.8 = 69 N, seven times the bag's weight of 9.8 N, so the net force is about 59 N upward; at equilibrium (stretch 9.8 ÷ 85.75 = 0.11 m) the elastic force equals the weight. A cord of only 24.5 N/m, the value the shortcut would give, would stretch about 1.7 m, putting the lowest point 3.7 m below the platform, and the bag would hit the floor. A stiffer cord stops the bag higher, and a softer cord stops it lower: with this 2.0 m cord, 100 N/m gives a stretch of 0.73 m (lowest point 0.27 m above the floor) and 70 N/m gives 0.90 m (0.10 m above the floor). Only a cord softer than about 59 N/m lets the bag reach the floor, because a stretch of 1.0 m needs 9.8 × 3.0 = ½k × 1.0², so k = 58.8 N/m. Test, compare elastic and gravitational forces from the measured lowest point, and modify.

Investigation: find k and predict the bungee. Materials: a spring or rubber cord, a stand and clamp, slotted masses from 0.050 to 0.500 kg, a meter stick, safety goggles, and a floor mat. Steps: measure the relaxed length; add masses one at a time, letting each settle, and record the stretch; remove them to check the spring returns to its relaxed length (if not, you passed the elastic limit and that data point is discarded). Plot force vs. stretch and take the slope for k. Then hold a chosen mass at the relaxed length, release it, and mark the lowest point with a second student watching the meter stick; compare with 2mg/k. Record: relaxed length, each mass, force, stretch, k, the predicted and measured lowest points. A result that supports the claim: a straight line through the origin and a measured drop close to 2mg/k (agreement within about 10 percent is a suggested classroom target; compare the difference with the uncertainty in k and in reading the lowest point). Safety: goggles, because a slipping spring or snapping cord whips; masses within the spring's range; nobody's feet under the load. What the model cannot show: air resistance and internal friction in the spring make the real drop slightly shorter than 2mg/k, and a rubber cord may not be linear over the whole range, so the straight-line model is checked, not assumed.

Worked example 1 — spring constant from one measurement. A spring stretches 0.060 m when a 0.300 kg mass hangs at rest from it. The elastic force equals the weight, 0.300 × 9.8 = 2.94 N, so k = F/Δx = 2.94 ÷ 0.060 = 49 N/m. With a 0.500 kg mass the stretch would be 4.9 ÷ 49 = 0.10 m, provided the spring is still within its linear range.

Worked example 2 — compression. A spring with k = 800 N/m is compressed 0.050 m by a box resting on it on a table. The spring pushes up with 800 × 0.050 = 40 N, directed toward its relaxed position. If the box is at rest, its weight is 40 N and its mass is 40 ÷ 9.8 = 4.1 kg. The same k applies to compression as to stretch.

Worked example 3 — net force at the lowest point. A 0.40 kg mass on a 25 N/m spring is pulled to a stretch of 0.30 m and released. Spring force 25 × 0.30 = 7.5 N up; weight 3.92 N down; net 3.58 N up; acceleration 3.58 ÷ 0.40 = 9.0 m/s² upward at that instant. The equilibrium stretch is 3.92 ÷ 25 = 0.157 m, so the mass was released 0.14 m below equilibrium and will rise to about 0.14 m above it, a stretch of only about 0.01 m, before turning back (ignoring friction and air). The calculation here is Newton's second law at one instant, with the elastic force as one of the forces; the rest of the motion is described, not calculated, and the turning point is a symmetry you can check with a meter stick.

Worked example 4 — reading a force-stretch graph with an offset. Data for a spring: force 1.0, 2.0, 3.0 and 4.0 N give stretches 0.00, 0.020, 0.040 and 0.060 m. The line has slope 1.0 N ÷ 0.020 m = 50 N/m but does not pass through the origin: the first newton produced no stretch. The spring is pre-tensioned, its coils pressed together until about 1.0 N is applied, so Hooke's law applies to the extra force: F − 1.0 N = kΔx, with Δx measured from where stretching actually begins. The spring constant is still 50 N/m, but F itself is not proportional to the stretch.

Representation task: five free-body diagrams. For a 0.20 kg mass on a spring with k = 10 N/m, released from the relaxed length, draw the mass and its two forces (weight 1.96 N and spring force) at five positions: the release point (stretch 0), halfway to equilibrium (stretch 0.098 m), equilibrium (0.196 m), halfway to the lowest point (0.294 m) and the lowest point (0.392 m). Compute the spring force at each (0, 0.98, 1.96, 2.94, 3.92 N), draw the arrows to scale, and beside each write the net force with its direction and whether the mass is speeding up or slowing down there.

Independent performance: design the bungee. For a fragile object of your own choosing (a bag of flour or a raw egg in a bag), a platform height set by your teacher, and an available cord whose k you must measure at the length you will use (cutting a uniform cord to half its length roughly doubles its k, so recalibrate after changing the length), calculate the cord length and predicted lowest point so the object stops within a specified distance of the floor. Build, test and revise. Success criteria: the force-stretch graph for the cord is shown with its slope and its range of validity; the prediction is written before the test; the measured lowest point is compared with the prediction; the elastic and gravitational forces on the object at the equilibrium stretch and at the lowest point are computed and compared; and the design revision states which quantity was changed and why.