Conservation of Energy Formula: K₁ + U₁ = K₂ + U₂, Examples

The conservation of energy formula for mechanics is $\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2$ : when only gravity (or another conservative force such as a spring) does work, kinetic energy plus potential energy at the start equals kinetic energy plus potential energy at the end. In shorthand, $K_1 + U_1 = K_2 + U_2$ .

The general law is wider: energy cannot be created or destroyed, only changed from one form to another, so the total energy of an isolated system stays constant. When friction or air resistance acts, some mechanical energy turns into heat, and the formula gains a term for that loss: $K_1 + U_1 = K_2 + U_2 + E_{\text{lost}}$ . This page shows how to use both versions, with worked problems on roller coasters, pendulums, springs and sledges, and a calculator that finds speed or height drop.

Quick facts

Mechanical form
$K_1 + U_1 = K_2 + U_2$
With gravity
$\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2$
Speed after a drop h
$v = \sqrt{u^2 + 2gh}$ ; mass cancels
With friction
$K_1 + U_1 = K_2 + U_2 + |W_f|$
Unit of each term
joule (J)
Holds when
Only conservative forces do work, for the mechanical form
Courses
Class 9, Class 11, AP Physics 1, JEE

Conservation of energy formula physics students use

Terms in the energy balance
Term	Meaning	Unit
$\frac{1}{2}mv_1^2$	Energy of motion before	J
$mgh_1$	Stored energy of height before	J
$\frac{1}{2}mv_2^2$	Energy of motion after	J
$mgh_2$	Stored energy of height after	J
$\frac{1}{2}kx^2$	Spring energy, added when a spring is involved	J
$E_{\text{lost}}$	Energy turned into heat and sound by friction	J

Every term contains m except the spring term, so in pure gravity problems the mass cancels. Divide through by m and you get $\frac{1}{2}v_1^2 + gh_1 = \frac{1}{2}v_2^2 + gh_2$ . That is why a heavy and a light ball dropped from the same height reach the ground at the same speed, if air resistance is small.

How to use the formula for conservation of energy

Pick two moments, the start and the end, and sketch the object at each.
Choose a zero level for height. The lowest point in the problem is usually easiest.
List the forms of energy present at each moment: kinetic, gravitational, spring.
Decide whether friction or drag acts. If it does, include the energy lost.
Write the balance and solve for the unknown. Check that the answer is sensible.

This page's solver covers the case you meet most: an object with speed u falls through a height h, and you want its new speed v, using $v = \sqrt{u^2 + 2gh}$ with g = 9.8 m/s². Enter a negative h for a rise.

Worked examples

Example 1: a roller coaster

A coaster car is let go, motionless, on the crest of a 20 m hill. Ignoring friction, find its speed at a point 5.0 m above the ground.

Height drop h = 20 − 5.0 = 15 m, and u = 0.

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 15} = \sqrt{294} = 17\ \text{m/s}$

The mass of the car is never needed.

Example 2: a pendulum bob

A pendulum bob is pulled aside so it is 0.20 m above its lowest point, then released. Find its speed at the bottom.

$v = \sqrt{2 \times 9.8 \times 0.20} = \sqrt{3.92} = 1.98\ \text{m/s} \approx 2.0\ \text{m/s}$

Example 3: a spring launcher

A launcher spring (stiffness 200 N/m) is squashed by 10 cm and releases a 0.50 kg block on a smooth level table. Find the block's launch speed.

Convert: x = 0.10 m. Spring energy turns into kinetic energy:

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 \Rightarrow \frac{1}{2}(200)(0.10)^2 = 1.0\ \text{J}$

$v = \sqrt{\frac{2 \times 1.0}{0.50}} = \sqrt{4.0} = 2.0\ \text{m/s}$

Example 4: a sledge with friction

Sitting on a sledge, a 40 kg child slides off a 10 m high bank from rest and is moving at 12 m/s at its foot. How much energy did friction take?

$U_1 = mgh = 40 \times 9.8 \times 10 = 3920\ \text{J}$

$K_2 = \frac{1}{2}(40)(12)^2 = 2880\ \text{J}$

$E_{\text{lost}} = 3920 - 2880 = 1040\ \text{J} \approx 1.0\ \text{kJ}$

Without friction the speed would have been √(2 × 9.8 × 10) = 14 m/s.

Why the law holds and its link to work

Start from the work energy theorem: net work done on a body shows up as a gain or loss of its kinetic energy. Gravity does work $mg(h_1 - h_2)$ as an object falls, so $mg(h_1 - h_2) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$ . Move the terms around and you have the conservation equation. Potential energy is simply a way of book keeping the work a conservative force will do.

Friction is not conservative, because the work it does depends on the path. That work, from the work done by friction page, shows up as the energy lost term. In collisions, kinetic energy is often lost too; the coefficient of restitution measures how much. Total energy, counting heat and sound, is still conserved.

Mistakes that cost marks

Using the mechanical form when friction acts. Read the question for words such as rough, drag or resistance.
Measuring heights from different zero levels at the start and end.
Forgetting that a falling body's potential energy decreases, so the height drop goes on the kinetic side.
Using the vertical drop wrongly on a slope. Only the change in height matters, not the length of the slope.
Giving an energy a direction. Joules simply add as numbers.

For each form of energy in the balance, see the kinetic energy, potential energy and formula of energy pages; the work and energy hub lists every other page in the chapter.

Frequently asked questions

What is the conservation of energy formula?

In mechanics it is K₁ + U₁ = K₂ + U₂, or ½mv₁² + mgh₁ = ½mv₂² + mgh₂ near Earth. In general, the total energy of an isolated system stays the same.

Does the law of conservation of energy hold when there is friction?

Total energy is still conserved, but mechanical energy is not. Friction changes some kinetic or potential energy into heat, so you add that lost energy to the final side.

Why does mass cancel in conservation of energy problems?

When gravity is the only force doing work, every term in the energy balance has a factor m, so you can divide it out. The speed then depends only on the height drop.

What is the speed of an object after falling through a height h?

From energy conservation, v = √(u² + 2gh). Starting from rest, v = √(2gh).

Is energy conserved in a collision?

Total energy always is. Kinetic energy is conserved only in perfectly elastic collisions; in others, some becomes heat, sound or deformation.