# Conservation of energy of a jet powered car

1. Dec 16, 2008

### nns91

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

1. A block mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case ??

2. When the jet powered car went out of control during a test drive, it left skid marks about 9.5km long. (a) If the car was moving initially at a speed of v=708km/h, estimate the coefficient of kinetic friction. (b) what was the kinetic energy K of the car at time t=60s after the brakes were applied ? Take the mass of the car to be 1250kg

2. Relevant equations

E=PE +KE +Uel

3. The attempt at a solution

1.I did like this:

Initial: E=kx^2/ 2

Final:E=mv^2/2

Then I set them equal. and solve for x. When I did the final energy, I substitute m by 4, and v^2 by 9V^2

However I got a wrong answer. Where did I do wrong ???

2. I also did the same thing for this problem

Initial: E= mv^2/2

Final E=0

Emech= mv^2/2 = -Etherm

Then I set E therm =Fx=mg*coeff of kinetic friction*x

Then I solved for coeff by substituting x=9500m and m=1250kg.

I got coeff= 0.001 It was wrong. Can anyone check my answer ??

For part (b), I use vf^2= Vi^2 + 2ax to find a=-2.036

The v=v0 +at and get 74.54

KE= mv^2/2

I got 46.6 KJ

I was wrong again. Anyone know ??

2. Dec 16, 2008

### D H

Staff Emeritus
Show a bit more work, please. It's a bit hard to tell where you went wrong when you didn't supply enough information for us to deduce where that happened.

Same thing here: Show your work. In this case I suspect you have a units conversion problem.

3. Dec 16, 2008

### G01

For #1:

You forgot about friction. Remember that friction is removing energy from the system as the block slides.

For #2

Did you remember to convert the velocity to m/s?

4. Dec 16, 2008

### D H

Staff Emeritus
There is no friction in problem #1. The OP didn't tell us what answer he arrived at, merely that it was wrong.

That is what I suspect went wrong in this problem.

5. Dec 16, 2008

### nns91

I did convert to m/s as 196.7 m/s

Can you explain more about problem 1 ??

1. Here is my full work:

Initial: E=kx^2/ 2

Final:E=mv^2/2

Ef= Ei
kx^2/2=(4m*9v^2)/2
kx^2=4m*9v^2
x= squaroot( (4m*9v^2)/k )

2.

Initial: E= mv^2/2

Final E=0

Emech= mv^2/2 = -Etherm = (1250*196.7)/2 = 122916.7 J

Then I set E therm =Fx=mg*coeff of kinetic friction*x

122916.7 =1250* 9.81 * coeff *9500

Then I solved for coeff by substituting x=9500m and m=1250kg.

I got coeff= 0.001

For part (b), I use vf^2= Vi^2 + 2ax to find a

0=196.7^2 +2a*9500
a=-2.036 m/s^2

The v=v0 +at = 196.7 -2.036*60
v=74.54 m/s

KE= mv^2/2
KE=1250*74.54 / 2

I got 46.6 KJ

6. Dec 16, 2008

### G01

Apparently I read "frictionless" and it translated as "with friction." Oops. Sorry about that.

For #1:

That seems to be the correct compression, but maybe they want the second compression, call it $x_2$, in terms of the first compression, $x_1$.

i.e. Do they want the compression of the spring with the "4m" block in terms of the compression of the spring with the "1m" block?

For #2:

When you found the kinetic energy you forgot to square the speed.

Last edited: Dec 16, 2008
7. Dec 16, 2008

### nns91

They do want the 2nd compression. How will it be different ?

8. Dec 16, 2008

### G01

Using the wrong kinetic energy will change the value you compute for the coefficient. Have you tried to find the coefficient now that you know your mistake with the kinetic energy?

For the first one:

What I am trying to say is that they may want your answer in the following form:

(compression with second block) = (some constant)*(compression with first block)

9. Dec 16, 2008

### nns91

For number one , I think so.

10. Dec 16, 2008

### nns91

I think the problem with my first problem here is I have to substitute K somehow. Since K is not give so I cannot give my answer in terms of k.

11. Dec 17, 2008

### D H

Staff Emeritus
That's correct. To do that you need to use the information provided regarding the reaction of the first block, and you didn't do that.

12. Dec 17, 2008

### nns91

So you mean, I have to calculate K from the first one ??

13. Dec 17, 2008

### D H

Staff Emeritus
That's the basic idea. Try it.

14. Dec 17, 2008

### nns91

I got k=mv^2/x^2.

Then I did the subsitution and everything kinda cancel out

15. Dec 17, 2008

### D H

Staff Emeritus
You should get something along the lines of what G01 alluded to in post #8.