Does it matter if the spring is stretched or compressed?

No, because the displacement $x$ is squared in the elastic potential energy formula. Whether $x$ is positive or negative, squaring it makes the energy positive. The spring stores the same amount of energy whether you stretch it $10 \text{ cm}$ or compress it $10 \text{ cm}$.

What if the spring is hanging vertically?

If the spring is vertical, you must also include gravitational potential energy ($U_g = mgh$) in your conservation of energy equation. As the mass moves up and down, it exchanges energy between kinetic, gravitational potential, and elastic potential forms.

How do I find the spring constant if it isn't given?

Sometimes you have to use Hooke's Law ($F = kx$) first. If a problem tells you how much force is required to stretch the spring a certain distance, you can use $k = \frac{F}{x}$ to find the spring constant before you start the energy part of the problem.

How do you solve a problem with springs and energy?

Solving a problem with springs and energy involves using the principle of conservation of energy. You compare the total energy of a system at two different points in time, knowing that the energy just changes forms, like from motion to stored spring energy.

Think of a spring like a bank account. You can deposit energy into it by pushing or pulling (which stores elastic potential energy), and you can withdraw that energy to make something move (turning it into kinetic energy). The total amount of money (energy) stays the same if there are no outside fees, like friction.

What is Elastic Potential Energy?

When you stretch or compress a spring, it stores energy. This is called elastic potential energy. The formula is $U_s = \frac{1}{2}kx^2$ . Here, $k$ is the spring constant, which tells you how stiff the spring is. A higher $k$ means a stiffer spring. The $x$ represents the displacement, which is how far the spring has been stretched or compressed from its natural, relaxed length.

Setting up the Conservation of Energy

To solve these problems, you write an equation that says the total initial energy equals the total final energy: $E_i = E_f$ . You look at the start of your problem and add up any kinetic energy (from movement), gravitational potential energy (from height), and elastic potential energy (from the spring). Then you do the same for the end of the problem. If a moving block hits a spring and stops, its initial kinetic energy completely transforms into final elastic potential energy.

Where Students Slip Up

A common mistake is confusing the total length of the spring with the displacement $x$ . The variable $x$ is only the change in length from the spring's normal, unstretched state. Another frequent slip is forgetting to square the $x$ or the $v$ in the energy equations. Always double-check your exponents!

Worked through

A $2 \text{ kg}$ block slides at $4 \text{ m/s}$ on a frictionless horizontal surface and hits a spring with a constant $k = 200 \text{ N/m}$ . How far does the spring compress before the block stops?

First, identify the initial and final states. Initially, the block is moving (kinetic energy) and the spring is relaxed (zero elastic potential energy). Finally, the block is stopped (zero kinetic energy) and the spring is compressed (elastic potential energy).

Set up the conservation of energy equation: $K_i = U_{sf}$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

Multiply both sides by 2 to cancel the fractions: $mv^2 = kx^2$

Now, solve for $x$ : $x^2 = \frac{m}{k}v^2$ $x = v \sqrt{\frac{m}{k}}$

Plug in the given numbers: $x = 4 \sqrt{\frac{2}{200}}$ $x = 4 \sqrt{0.01}$ $x = 4 (0.1) = 0.4 \text{ m}$

The spring compresses by $0.4$ meters.

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Where this comes from: OpenStax College Physics, Chapter 7: Work, Energy, and Energy Resources · Khan Academy: Work and energy unit · Fundamentals of Physics by Halliday, Resnick, and Walker