# Work-Kinetic Energy

1. Oct 24, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
With brakes fully applied, a 1540 kg car decelerates from a speed of 94.0 km/hr. What is the work done by the braking force in bringing the car to a stop? What is the change in the kinetic energy of the car?

2. Relevant equations

3. The attempt at a solution
W = -Fd
Not sure how to start this problem?

2. Oct 24, 2007

This problem is, I think, most easily solved by finding the initial kinetic energy of the car.
Calculations are simplified if you convert 94.0 km/hr to m/s ==> 94.0 km/hr ~ 26.111111 m/s.
Now, K = 1/2 m v^2 , so the initial kinetic energy of the car is given by K = (1/2)x(1540 kg)x(26.111111 m/s)^2 = approx. 525 kJ.
Since the final velocity of the car is zero, the final kinetic energy is zero as well. Thus, the change in kinetic energy is 0 - 525 kJ = -525 kJ.
The work is equal to the total change in energy, so W = -525 kJ as well.
===> change in KE = W = -525 kJ <===

Last edited: Oct 24, 2007
3. Oct 24, 2007

### BuBbLeS01

I understand...but its saying its incorrect?

4. Oct 24, 2007

well, i may have rounded a bit too much... also, it's common just to write the magnitude of the work, so try W = 525 kJ if -525 kJ doesn't work. A more exact answer is 524978.3951 J

5. Oct 24, 2007

### BuBbLeS01

I got the change in kinetic energy right but, is the work is wrong when entered as positive. So do I enter is negative?

6. Oct 24, 2007
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