Conservation of Energy

You might remember from chemistry the law of conservation of mass: Mass is neither created nor destroyed, it only changes forms… (If you were thinking about discussing the equation E = mc2 at this point – Please don’t. We will discuss it later. We have to learn some lesser laws before we can discuss the granddaddy of them all!) Well there is also a law called the conservation of energy: Energy is neither created nor destroyed, it only changes forms. The rule to remember is that every kind of energy can be transferred into another kind of energy. Of course since this is physics there is an equation (or several) to describe this situation.

E = 0 or E0 = E

E is energy measured in Joules (of course)

Of course this law can sometimes be a little too simple to be mathematically practical so of course there are several other equations. Here is the big one that should work in every situation

Wdone + U0 + K0 = U + K + Wfriction

Translated into words this says the Work done during the process plus the initial potential energy plus the initial kinetic energy equals the final potential energy plus the final kinetic energy plus the work lost due to friction. The frictional work is usually lost to the surroundings as heat!

This is a nice general equation that should work for all situations, but it is pretty big. Let’s look at what it breaks down to for some specific situations.

Example problem 1: A 3 kg toy car is moving at 2 m/s has a forward pull of 7 N exerted on it for 1.5 m. What is the final velocity of the car?

In this case height of the car is never mentioned and friction/energy lost is not mentioned and therefore potential energy (U) and Wfriction don’t need to be considered. That is how you use the big equation! You cross out or take out what you don’t need. So here is what we are left with.

Wdone + K0 = K

Substituting in the equations for W and K

Fx + ½ mv02 = ½ mv2

The above equation rearranged would be

W = K-K0 or W = K

which is the work energy theorem.

Example problem 2: A set of keys is dropped from the top of the TaosGorgeBridge (height 650 ft). What is the speed of the keys as they hit the river?

You remember from kinematics that dropped means v0 = 0 so there is no initial kinetic energy. No work or friction is mentioned so all the W’s get left off. There is no height left at the end so we are left with:

U0 = K

Substituting:

, mgh0 = ½ mv2

dividing both sides by m and rearranging we get

v =

This is a very famous equation and makes the solution to our problem easy.

Example 3. A set of twins is riding a sled (mass of twins and sled = 40 kg) down a hill and there is no friction in the snow. The twins’ older brother gives them a running start at the top of the hill of 4 m/s. The vertical height of the hill is 12 m. The angle of the hill is 30. At the bottom of the h ill they travel 15 m along level ground before starting up another hill. The angle of the next hill is 40. What is the velocity of the twins and sled at a vertical height of 5 m on the next hill? The picture below is a representation of the path the sled follows. The box is the sled.

Example 4. A set of twins is riding a sled (mass of twins and sled = 40 kg) down a hill and there is friction. The twins’ older brother gives them a running start at the top of the hill of 4 m/s. The vertical height of the hill is 12 m. The angle of the hill is 30. At the bottom of the hill they travel 15 m along level ground before starting up another hill. The angle of the next hill is 40. The velocity of the twins is 10 m/s at a vertical height of 5 m on the next hill. What was the energy loss due to friction? Use the same picture as before.