# Determine the energy required to accelerate a car

1. Feb 7, 2006

### Pepsi24chevy

energy problem....

I got a problem that goes likethis: Determine the energy required to accelerate a 1300 kg car from 10 to 60km/h on an uphill road with a vertical rise of 40m.

Ok i know that work is the itegral of force which is ma. But what about this vertical rise of 40m? Does this basically mean at a 90 degree angle? Guess i just don't see the problem happening like that, but if that is what it means, that is what it means.

2. Feb 7, 2006

### Staff: Mentor

A simpler way to approach this problem would be to look at the change in kinetic and potential energy of the car. What are the starting KE and PE, and what are the ending KE and PE? What is the total energy of the car at the start and end?

3. Feb 7, 2006

### Pepsi24chevy

So i got 1/2*1300[(60000/2600)^2-(10000/3600)^2]*((1kJ/kg)/(1000m^2/s^2)) for the change in the kinetic energy. I still don't see where the vertical rise of 40m comes into play...

4. Feb 7, 2006

### Staff: Mentor

Think about what happens when you throw a ball straight up in the air. When it leaves your hand, it has some initial velocity, and hence some Kinetic Energy (KE). At the top of its arc, its velocity is zero, so what is the KE of the ball then? Where did that energy go? How come the faster you throw the ball, the higher the top of the arc is? What is the relationship between the initial velocity and the distance to the top of the arc?

The same concepts apply to the problem that you are trying to solve here.

5. Feb 7, 2006

### Cyrus

Hey, are you in my thermo class? That is due tomorrow. Look at the example problem in the book, it already shows you how to do it. Open your book to page 69 example 2-8 and 2-9.

6. Feb 7, 2006

### assyrian_77

Think of it as a slope. The car starts at the bottom (h=0) and "finishes" at the top of the slope, 40m above starting ground level. The angle is not important.

As posted earlier, thinking about the energies at the start and at the end, you should be able to solve the problem.

7. Feb 7, 2006

### Pepsi24chevy

ok thanks, I got it now. I wasn't clearly thinkin about how the 40 factors in.

8. Feb 7, 2006

### Cyrus

Are you taking thermo with jackson?

That is not true. Work is the integral of $$\int_c F * ds$$

Last edited: Feb 7, 2006