Conservation of energy of a jet powered car

[nns91]

Homework Statement

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

Homework Equations

E=PE +KE +Uel

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

E_mech= mv^2/2 = -E_therm

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 ??

 

[D H]

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

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.

2. I also did the same thing for this problem

Initial: E= mv^2/2

Final E=0

E_mech= mv^2/2 = -E_therm

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 ??

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

 

[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?

 

[D H]

For #1: You forgot about friction.

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

For #2: Did you remember to convert the velocity to m/s?

That is what I suspect went wrong in this problem.

 

[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

 

[G01]

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.

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

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

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.

 

[nns91]

How about the coefficient ??

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

 

[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)

 

[nns91]

For number one , I think so.

 

[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.

 

[D H]

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.

 

[nns91]

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

 

[D H]

That's the basic idea. Try it.

 

[nns91]

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

Then I did the subsitution and everything kinda cancel out

 

[D H]

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