Friction (Sliding)

[Kaln0s]

Homework Statement

I'm positive Part 1 is right, 2 and 3 I'm sketchy with.

1. A race car with mass 1000 kg moves at speed 40 m/s. What is the kinetic energy of the car in kiloJoules? Do not write the unit in your answer.

For this I got 1/2(1000)(40^2) = 800000 convert to kilojoules = 80

2. The car in the previous problem locks its brakes and slides to a stop due to a constant frictional force. How much work was done by this force? Express your answer in kiloJoules but as usual do not write the unit as part of your answer.

= 1/2(1000)(40^2) - 1/2(1000)(0^2) = 80000 = 80kilojoules?

3. If the car in the previous two problems slides to rest with a constant frictional force of magnitude 10,000 N, how far does it slide? Express your answers in meters, but do not write the unit in your answer.

This one is below.

Homework Equations

1/2m(v^2)
vf^2 = vi^2 - 2*a*d

The Attempt at a Solution

I'm really not sure how to do this. I tried finding acceleration which I believe is 10,000N / 1000kg = 10m/s

Then I just tried to use an equation of kinematics...

= 0^2 = 25^2 + 2(10)(d) and I get -31.25 for that, but I don't think that's right ^_^;.

I appreciate any help

 

[Delphi51]

1 and 2 look good.
For 3 you have
vf^2 = vi^2 - 2*a*d
0^2 = 25^2 + 2(10)(d)
Isn't that formula supposed to have a plus sign instead of a minus?
Of course when you put in a = -10, you'll then get a minus sign.
And the 25 should be 40.

Also can be done using the work formula - a little less "work" involved.

 

[Kaln0s]

1 and 2 look good.
For 3 you have
vf^2 = vi^2 - 2*a*d
0^2 = 25^2 + 2(10)(d)
Isn't that formula supposed to have a plus sign instead of a minus?
Of course when you put in a = -10, you'll then get a minus sign.
And the 25 should be 40.

Also can be done using the work formula - a little less "work" involved.

Why in the hell did I plug in 25? I kill myself sometimes...

so yeah 1600 = 20d ----> d = 80

Thank you very much sir.