Chemical Potential Energy and Kinetic Energy

[pewpew23]

Homework Statement

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

Homework Equations

KE = 1/2mv^2

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.

 

[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).

 

[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?

 

[Matterwave]

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

KE final - KE initial = energy required

 

[pewpew23]

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

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

512m/1536 = 3 times as much

 

[Matterwave]

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