### Kinetic Energy 2

Change in Kinetic Energy

Example 1
Assume that the Ek of a car is at first 200J. And after a while the Ek is 300J. Calculate the work done on the car.

Since the Ek of the car increased, we conclude that work must have been done on the car.
The amount of work done is:
Work Done = Ek(final) – Ek(initial)
Work Done = 300 – 200
Work Done = 100J

This is often called ΔEk, the change in Ek.
So ΔEk is the work done on the car.
ΔEk = Work Done = Ek(f) – Ek(i)

Example 2
In this example, the Ek changed from 400J to 100J. Calculate the work done on the car.

ΔEk = Ek(f) – Ek(i)
=100 – 400
=-300J

The car lost 300J of kinetic energy. Why would the car lose energy?
1. The driver could have pressed the brakes. Then we say that 300J of work was done by the brakes in slowing down the car.

2. Possibly the road was rough. Then we can say that 300J of energy was used up in overcoming the friction surface of the road. We can even say that the road did 300J of work against the car!

Example 3
The Ek of the car at the bottom of the hill is 600J, and at the top the Ek is 200J.
Assume that the driver had not pressed the accelerator pedal. What did it “cost” to climb the hill?

Since the car lost 400J of Ek, we can say:

1. The hill “absorbed” 400J of the car’s Ek.
(By the way, the hill’s energy is called Gravitational Potential Energy – or Ep
So we can say Ep = 400J)

2. We can say the Work Done to climb the hill is 400J.

Extra Question
What if the driver wanted to still travel with 600J Ek in spite of the hill?
(Remember the hill would slow him down!)
How much extra energy must be produced by the engine to ensure this?

The engine would need to develop an additional 400J to “pay” for climbing the hill.
(Remember it costs 400J to climb the hill.)

The full energy developed by the engine would then need to be 1000J to achieve both requirements, i.e. to climb the hill AND keep its speed.
(600J to maintain the original speed and 400J to climb the hill)