Elastic potential energy of spring — Hooke's law applications and combinations

Question

A spring with spring constant $k = 200$ N/m is compressed by $0.1$ m. A block of mass $0.5$ kg is placed against it and released. Find the speed of the block when it leaves the spring on a frictionless surface. Also, if two such springs are connected in series, what is the effective spring constant?

(JEE Main & NEET standard problem)

Solution — Step by Step

By Hooke’s law, the force in a spring is $F = -kx$ , and the potential energy stored when compressed or stretched by $x$ is:

PE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = 1 \text{ J}

On a frictionless surface, all spring PE converts to KE when the block leaves the spring (at natural length):

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

1 = \frac{1}{2}(0.5)v^2

v^2 = 4 \implies v = \mathbf{2 \text{ m/s}}

For springs in series, the weaker combination rule applies:

\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{200} + \frac{1}{200} = \frac{1}{100}

k_{\text{eff}} = \mathbf{100 \text{ N/m}}

Series springs are softer — the effective $k$ is smaller than either individual spring.

Why This Works

A spring stores energy through deformation. When you compress it by $x$ , you do work against the restoring force. This work gets stored as elastic PE. Upon release, this PE converts back to KE — the spring pushes the block.

The energy stored is $\frac{1}{2}kx^2$ (not $kx^2$ ) because the force increases linearly from $0$ to $kx$ during compression — so the average force is $kx/2$ .

graph TD
    A["Spring Problem"] --> B{"What's asked?"}
    B -->|"Speed/velocity"| C["Energy Conservation<br/>½kx² = ½mv²"]
    B -->|"Effective k"| D{"Combination type?"}
    D -->|"Series"| E["1/k_eff = 1/k₁ + 1/k₂<br/>Result: softer"]
    D -->|"Parallel"| F["k_eff = k₁ + k₂<br/>Result: stiffer"]
    B -->|"Force at extension x"| G["F = kx<br/>Hooke's Law"]
    B -->|"Time period of SHM"| H["T = 2π√(m/k)"]

Alternative Method — Using Work Done by Spring Force

Instead of energy conservation, calculate work done by the spring directly:

W_{\text{spring}} = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 = \frac{1}{2}(200)(0.01) - 0 = 1 \text{ J}

By work-energy theorem: $W = \Delta KE$ , so $1 = \frac{1}{2}(0.5)v^2$ . Same result.

Remember the analogy: springs in series combine like resistors in parallel (reciprocal addition), and springs in parallel combine like resistors in series (direct addition). This is the opposite of resistors — don’t mix them up!

Common Mistake

Students often write PE $= kx^2$ instead of $\frac{1}{2}kx^2$ , forgetting the factor of $\frac{1}{2}$ . This gives a speed that’s $\sqrt{2}$ times too large. The factor comes from integration: $\int_0^x kx\,dx = \frac{1}{2}kx^2$ . If your answer seems unexpectedly large, check whether you missed this factor.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Work Done by Spring Force

Common Mistake

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