# Work done using time, velocity and mass?

1. Mar 30, 2016

### TDizzl

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

The law of conservation of energy applies to the motion of vehicles. Find the work done on a 1200kg vehicle when it slows from 90kmh-1 to 50kmh-1 in 8.0 seconds.
2. Relevant equations

W=1/2mv^2 - 1/2mu^2
3. The attempt at a solution

W=0.5*1200*50^2 - 0.5*1200*50^2

2. Mar 30, 2016

### cnstntcnfsn

If you assume a constant deceleration, you can use constant acceleration equations such as x = 1/2(vo + v)*t and v = vo + a*t. Work = Fd, where F is the force and d is the displacement. This can be expanded to W = m*a*d. Using the acceleration equations, you should be able to find all of your unknowns and plug them into the work equation.

3. Mar 30, 2016

### haruspex

Is that what you meant to post? What happened to the 90kmph?
Why would TDizzl care what the rate of deceleration is?

4. Mar 31, 2016

### TDizzl

Its W=0.5*1200*50^2 - 0.5*1200*90^2

5. Mar 31, 2016

### TDizzl

The question is from a paper which does not require the use of constant acceleration equations. It would seem bizarre to use a formula that hasn't been introduced to us yet.
Also, the solution is written as -2.59*10^5 Joules in the paper.

6. Mar 31, 2016

### PeroK

7. Mar 31, 2016

### TDizzl

No, that was how I substituted the values in as an attempt to answer the question.

8. Mar 31, 2016

### haruspex

In what units?

9. Mar 31, 2016

### TDizzl

m=1200kg
v=50kmh-1
u=90kmh-1

10. Mar 31, 2016

### haruspex

So what is it in Joules?

11. Mar 31, 2016

### TDizzl

Yea its like -3360000J, that's what I got

12. Mar 31, 2016

### haruspex

No, the speeds you are given are in kmh-1, not ms-1.

13. Mar 31, 2016

### TDizzl

So how do I get the answer as -2.59×105?

14. Mar 31, 2016

### haruspex

How do you convert km/h to m/s?

15. Mar 31, 2016

### TDizzl

Alright -2.59×105 is the answer I got, the problem was that I was doing 502/3.6 instead of (50/3.6)2.
Thanks for the help, feeling extremely stupid.