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A block slides into a spring, squashes it and bounces back. A ball drops onto a spring and keeps going after first touching it. These questions look like force problems, but the spring force changes at every point, so using F = ma directly needs calculus. Energy avoids that. This post covers the three questions you will be asked about any spring-block system, with worked examples for each.
Where does ½kx² come from?
An ideal spring obeys Hooke's law, F = −kx, where x is the displacement from the natural length and k is the spring constant in N/m. NCERT's Work, Energy and Power chapter integrates this force to get the work done by the spring when it is stretched from 0 to x: W = −½kx². The stored elastic potential energy is therefore U = ½kx², the same whether the spring is stretched or compressed by x.
You can also read it off the force-extension graph: the force rises in a straight line from 0 to kx, so the area under it is a triangle, ½ × x × kx. The JEE (Main) 2026 syllabus lists "the potential energy of a spring" and conservation of mechanical energy in its Work, Energy and Power unit.
OpenStax University Physics adds a point that saves confusion later: potential energy belongs to the system and its interaction, here the block and the spring, not to the block alone.
How do you find the maximum compression?
At maximum compression the block stops for an instant, so all its kinetic energy has gone into the spring (on a smooth surface).
Example 1. A 2 kg block moving at 4 m/s on a smooth floor hits a spring with k = 800 N/m. Find the maximum compression.
Kinetic energy: ½ × 2 × 4² = 16 J.
At maximum compression: ½ × 800 × x² = 16, so x² = 0.04 and x = 0.2 m.
The shortcut x = v√(m/k) = 4 × √(2/800) = 4 × 0.05 = 0.2 m gives the same result.
Now add friction. Suppose the floor under the spring is rough, with μ = 0.25, so friction 0.25 × 2 × 10 = 5 N acts while the spring compresses (g = 10 m/s²). Friction takes energy 5x out of the system:
16 = ½ × 800 × x² + 5x, so 400x² + 5x − 16 = 0.
The positive root is x = (−5 + √25 625) / 800 ≈ (−5 + 160.08) / 800 ≈ 0.194 m. Slightly less than before, as it should be.
How fast does a spring launch a block?
Run the same idea backwards. A spring with k = 800 N/m is compressed 0.1 m and released, pushing a 0.5 kg block on a smooth floor. The stored energy is ½ × 800 × 0.1² = 4 J. All of it becomes kinetic energy: ½ × 0.5 × v² = 4, so v = 4 m/s.
The block leaves the spring at its natural length, because that is where the spring stops pushing.
Where is the speed greatest?
On a smooth horizontal floor, the block is fastest at the natural length, where the spring force is zero. On a vertical spring, gravity also acts, so the speed is greatest where the spring force equals the weight, at a compression of mg/k. Past that point the net force points up and the block slows down.
Example 2. A 1 kg ball is dropped from 0.4 m above the top of a vertical spring with k = 500 N/m. Take g = 10 m/s². Find the maximum compression and the maximum speed.
Maximum compression x: the ball falls a total of 0.4 + x before stopping. So 10 × (0.4 + x) = ½ × 500 × x², which gives 250x² − 10x − 4 = 0.
Positive root: x = (10 + √4100) / 500 ≈ (10 + 64.03) / 500 ≈ 0.148 m.
Maximum speed happens at compression x₀ = mg/k = 10/500 = 0.02 m.
Energy at that point: ½ × 1 × v² = 10 × 0.42 − ½ × 500 × 0.02² = 4.2 − 0.1 = 4.1 J, so v ≈ 2.86 m/s.
Compare this with the speed on first touching the spring, √(2 × 10 × 0.4) ≈ 2.83 m/s. The ball keeps speeding up for a short distance after contact, because the spring force is still smaller than its weight.
What mistakes do students make with springs?
Measuring x from the wrong point. x is always from the natural length, not from where the block started or from equilibrium.
Using kx² instead of ½kx², or using the spring force times distance as if the force were constant.
Assuming maximum speed at first contact on a vertical spring.
Mixing up two "extensions." A mass lowered slowly onto a hanging spring stretches it by mg/k. A mass attached at natural length and let go drops to 2mg/k, because nothing removes the energy.
Forgetting the change in gravitational energy when the spring is vertical or on an incline.
If the block oscillates after the collision, the time questions belong to simple harmonic motion; the energy method here still gives you the amplitude. For more practice with the energy equation itself, see our post on the work-energy theorem.
A practice routine for spring problems
Draw the spring at natural length, at the start and at the moment asked about. Label x at each.
Solve three horizontal problems: one smooth, one with friction, one where the spring launches the block.
Solve three vertical problems, each asking for maximum compression and maximum speed.
For every answer, check a limiting case: zero friction, zero drop height or a very stiff spring.
If you want to watch energy move between the block and the spring as it compresses, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, redo Example 2 with the drop height doubled and check that the maximum speed still occurs at 0.02 m compression.
