Friction Sliding Homework: Solve Part 1-3

In summary, the conversation involves solving three problems related to a race car with a mass of 1000 kg moving at a speed of 40 m/s. The first problem asks for the kinetic energy of the car, which is found to be 80 kilojoules. The second problem involves the car coming to a stop due to a frictional force, and the work done by this force is also found to be 80 kilojoules. The third problem asks for the distance the car slides with a frictional force of 10,000 N, which is found to be 80 meters using the formula for work.
  • #1
Kaln0s
6
0

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 :biggrin:
 
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  • #2
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.
 
  • #3
Delphi51 said:
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... :blushing:

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

Thank you very much sir. :approve:
 

Related to Friction Sliding Homework: Solve Part 1-3

1. What is friction and how does it affect sliding?

Friction is the force that resists the motion of two surfaces sliding against each other. It can affect sliding by either slowing down or preventing the movement of objects.

2. How is the coefficient of friction calculated?

The coefficient of friction is calculated by dividing the force of friction by the normal force. It is represented by the symbol "μ" and is a unitless value.

3. What factors can affect the coefficient of friction?

The coefficient of friction can be affected by various factors such as the type of surfaces in contact, the roughness of the surfaces, the weight of the objects, and the presence of lubricants or other substances between the surfaces.

4. How does the angle of inclination affect friction in sliding?

The angle of inclination, or the slope of the surface, can affect the coefficient of friction in sliding. As the angle increases, the normal force decreases, causing the coefficient of friction to decrease as well. This allows objects to slide more easily.

5. How can friction be reduced in sliding?

Friction can be reduced in sliding by using lubricants between the surfaces, using smoother surfaces, or reducing the weight of the objects. Additionally, changing the angle of inclination can also affect the amount of friction present.

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