Conservation of Energy

The law of conservation of energy states that the sum of all the energy in an isolated system remains the same over time. Energy can neither be added or subtracted. Energy can only be transferred from one form to another.

In an isolated system, the total energy at one point is the same as the energy at another point inside the system.

There are 3 main types of energy:

1. Kinetic energy: ½ * m * v2

2. Potential energy
i. gravitational: m * g * Δh
ii. spring/elastic: ½ * k * Δx2

3. Work of friction: – Ffriction * Δx = – (μ* Fnormal )* Δx

Therefore, the sum of the total energy, also called the mechanical energy or Emech is:

Emech = ΣKE + ΣPE

However, in the real world, resisting forces will act on objects in motion. Therefore, in these cases, the total energy at a particular point is:

Emech = ΣKE + ΣPE + Wfriction

Keep in mind that Wfriction is a negative value.

Recommended Steps

1. Isolate the problem in distinct checkpoints. Most problems will tell you to find the total energy from one point to another. Solve for the total energy at the first point and the second point separately. Keep in mind that at each point, the total energy is equal.
2. For situations in which 2 or more objects are present, find the energy of each object separately at each checkpoint.
3. Once you find the total energy at each point, equate them. Usually, a variable must be solved at one of the points.
4. Solution for the missing variable.

Practice Problems

1. A 10kg cart is being pushed along a hill with an initial velocity of 20m/s. The top of the hill is 20m high. Once the cart reaches the top, the person pushing the cart stops to take a break. However, the person accidentally bumps into the cart, and the cart rolls along the hillside to the bottom.

a. What is the total energy?

i. before ascending the hill?
ii. at the top of the hill?
iii. before reaching the ground again?

b. What is the velocity before the cart reaches the ground?
c. Was there energy loss occur during any part of the motion?


1. Divide the problem into distinct checkpoints.

The problem can be divided into three points: before the ascension, at the top of the hill, and before reaching the ground on the other side.

2. Find energy of each object separately at each checkpoint.

a. Before the ascension: there is no potential energy (assuming that the bottom of the hill is the zero marker). However, there is kinetic energy because the object is moving. Therefore, the total energy is

Emech = PE + KE = 0 + ½ * 10 * 20 ^2 = 4000 J

b. At the top of the hill: The cart is motionless, so there is no kinetic energy. However, the cart is 20m high from the datum, so there is gravitational potential energy. The total energy at this point is:

Emech = PE + KE = 10 * 9.81 * 20 + 0 = 1962 J

Notice that the total energy at point a is greater than the energy at point b. That means that there was an energy loss during the transition. The total energy loss from point a to b is:

4000 – 1962 = 2038 J

This represents the work performed from resisting forces (air drag, friction from hillside, etc.)

c. Before reaching the ground: since the object is on the ground again, there is no potential energy. However, there is a kinetic energy because the cart is in motion again.
Therefore, the total energy is:

Emech = PE + KE = 0 + ½ * 10 * Vfinal2

The final velocity must be solved.

3. Equate the final energies together and solve for missing variable.

You can use either point a or b to equate to c. Let’s equate the energies at point b and c:

1962 = ½ * 10 * Vfinal2

Solving for the final velocity,

Vfinal = 19.802 m/s

Notice that the initial and final velocity are almost the same.