# Homework Help: Silly work done on a car question.

1. Nov 30, 2011

### MarcZZ

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

It takes 139 kJ of work to accelerate a car from 22.8 m/s to 27.7 m/s. What is the car's mass?

2. Relevant equations

Well, we know that work is also known as J and J is = to kg*m^2/s^2 so we can use the equation J = m(vf - vi) to solve this I think.

3. The attempt at a solution

So We know that 139 kJ is equal to 139x10^3 J. Also m(27.7 m/s - 22.8 m/s) = m(4.9m/s), so combing these we assume that 139000 J = ?kg(4.9m/s) But from here I'm getting confused badly. I think I'm misusing the equation somehow as I'm not sure how to manipulate the 4.9m/s in order to get rid of the kg*m^2/s^2 units on the other side.

Any help will be much appreciated. :)

2. Nov 30, 2011

### Staff: Mentor

The work done will equal the change in the car's kinetic energy--not momentum. (That's why your units are not working out.) How would you calculate that?

3. Nov 30, 2011

### MarcZZ

KE = 1/2mv^2. I had pondered this, so in other words it's possible to use my change in velocity in place of V in the KE formula?

4. Nov 30, 2011

### Staff: Mentor

What you need is ΔKE = KEf - KEi.

5. Nov 30, 2011

### MarcZZ

I'm sorry but how can that be usable if you don't have the mass? Isn't it essentially equal to m = 2K / (delta V)^2. Sorry, I'm getting confused I just don't see this.

Edit: Never mind. I think I'm misunderstanding stuff because some notes I took in my lecture I miscopied. :/

6. Nov 30, 2011

### Staff: Mentor

You'll be using this to solve for the mass. (As you've probably figured out by now.)

7. Nov 30, 2011

### MarcZZ

Ahh never mind. I figure out what the problem was, I had used the formula you mentioned before but divided it by 2 as opposed to multiplying it. I've got it now the answer is 1.123x10^3 kg. Thanks very much for your help. :)