The spring and the conservation of mechanical energy

The energy account balances

Drop a ball. Gravity's work increases its kinetic energy and, by the same amount, decreases its potential energy. So the total stays the same:

This is the conservation of mechanical energy. It holds whenever only conservative forces do work. Forces that do no work, like the normal reaction on a smooth track, do not spoil it.

A mass on a spring

A block on a smooth floor is tied to a spring, pulled out by $A$ and let go. Its energy is

At $x = \pm A$ it stops for an instant: these are the turning points. At $x = 0$ all the energy is kinetic, and the block is fastest. Watch the energy move.

You should find the point sliding on the parabola between the turning points, and the kinetic energy (the gap up to the dashed line) largest at the middle.

Watch out. Energy is a scalar. Conservation of energy gives a speed, not a direction, and it says nothing about how long a motion takes.

A worked example

A $1.5\ \text{kg}$ box slides across smooth ice at $3.0\ \text{m/s}$ into a buffer spring with $k = 1350\ \text{N/m}$.

1. Largest squeeze. $\tfrac12 mv^2 = \tfrac12 kx^2$, so $x = v\sqrt{\dfrac{m}{k}} = 3.0\ \text{m/s} \times \sqrt{\dfrac{1.5\ \text{kg}}{1350\ \text{N/m}}} = 0.10\ \text{m}$.
2. Speed at a squeeze of $0.050\ \text{m}$. $\tfrac12 \times 1.5\ \text{kg} \times v^2 = \tfrac12 \times 1.5\ \text{kg} \times (3.0\ \text{m/s})^2 - \tfrac12 \times 1350\ \text{N/m} \times (0.050\ \text{m})^2 = 6.75\ \text{J} - 1.69\ \text{J} = 5.06\ \text{J}$.
3. So $v = \sqrt{\dfrac{2 \times 5.06\ \text{J}}{1.5\ \text{kg}}} = 2.60\ \text{m/s}$. Half the squeeze has taken only a quarter of the energy.

Think first. Write the energy at two moments, set them equal, and only then put the numbers in.

Practise with new numbers

Try it in Python

The program follows a falling ball and prints its potential, kinetic and total energy at five heights.

Remember these

• With only conservative forces working: $K_1 + U_1 = K_2 + U_2$.
• A spring: $\tfrac12 mv^2 + \tfrac12 kx^2 = \tfrac12 kA^2$; turning points at $x = \pm A$.
• Energy gives speeds, not directions or times.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A ball is dropped from $6.0\ \text{m}$. How fast is it moving $1.0\ \text{m}$ above the ground? ($g = 9.8\ \text{m/s}^2$.)

2. A mass on a spring with $k = 400\ \text{N/m}$ has $2.0\ \text{J}$ of energy in all. Where are its turning points?

3. Where in its motion is a mass on a spring moving fastest?

4. A pendulum on a smooth pivot swings back to the same height on the other side. Why?