# Chemical Potential Energy and Kinetic Energy

1. Jan 10, 2010

### pewpew23

1. The problem statement, all variables and given/known data

The chemical potential energy in a certain amount of gasoline is converted to kinetic energy in a car that increased its speed from 0 mph to 32 mph. The car accelerated to 64 mph. Compared to the energy required to go from 0 to 32 mph, the energy required to go from 32 mph to 64 mph is

(A) half as much
(B) as much
(C) twice as much
(D) three times as much
(E) four times as much

2. Relevant equations

KE = 1/2mv^2

3. The attempt at a solution

Is there any equation for chemical potential energy?
I don't know what to do with just the KE equation.

2. Jan 10, 2010

### Matterwave

The chemical potential energy in this question is a red herring. All you need is the kinetic energy. If you need twice as much kinetic energy, then you need twice as much chemical potential energy, it's a 1-1 correspondence (at least for simple problems like this, other factors such as engine efficiency at different temperatures are too complicated for this).

3. Jan 10, 2010

### pewpew23

so going from 0 to 64 would require twice as much kinetic and chemical potential energy as going from 0 to 32?

and going to 0 to 32 would require the same amount as going to 32 to 64?

4. Jan 10, 2010

### Matterwave

No, use your formula. KE=1/2mv^2

KE final - KE initial = energy required

5. Jan 10, 2010

### pewpew23

1/2m(32)^2 - 1/2(0)^2 = 512m

1/2m(64)^2 - 512m = 1536m

512m/1536 = 3 times as much

6. Jan 10, 2010

### Matterwave

Yea, looks good to me, except energy is expressed in J=joules not meters. =)