Dynamic question with a sliding car, looking for distance and power

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Homework Help Overview

The discussion revolves around a physics problem involving a car with a mass of 2250 kg that slides to a stop after locking its brakes, transitioning from a speed of 98 km/h to 80 km/h. Participants are tasked with calculating the distance traveled during the slide and the power generated by friction, given a coefficient of kinetic friction of 0.3.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the concept of treating the car as a block to simplify calculations related to friction and deceleration. Questions arise regarding the application of the coefficient of friction and its role in determining acceleration and distance. Some participants suggest using the work-energy theorem and relate it to the forces acting on the car.

Discussion Status

The discussion is active, with participants providing various insights and approaches to the problem. Some guidance has been offered regarding the use of friction and acceleration, while others express confusion about the calculations and the application of formulas. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants note the importance of converting units from km/h to m/s and discuss the implications of treating the car as a single block versus considering individual tires. There are also references to potential errors in calculations and the need for clarity in the application of physics principles.

Femme_physics
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Homework Statement



http://img263.imageshack.us/img263/1961/carslide.jpg

In a car whose mass (m) - 2250 kg, and drives in a velocity of 98 km/h, the driver activates the breaks and causes the four wheels to lock, and the sliding motion of a car through distance d, till decreasing the velocity to 80 km/h.

If the kinetic friction coeffecient is 0.3, calculate:

A) The distance (d) that the car traveled in the sliding
B) The power P (in kilowatts) that the friction coeffecient has developed during sliding.

Comment: 1 kW = 1 kN x m/s

The Attempt at a Solution

This is how I get stuck

http://img855.imageshack.us/img855/8250/theattempt1231231231232.jpg
 
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You don't really need to consider individual tires. Just pretend it's a block that is slowing down due to friction. What is the frictional force if you know the coefficient of kinetic friction and the weight of the car?
 
You don't really need to consider individual tires.

Because it's a simple question, but in reality you do have to consider them individually?
 
Femme_physics said:
Because it's a simple question, but in reality you do have to consider them individually?

In this case it doesn't matter.

Part of the weight may be on the front, and part on the back.
But if you multiply by the friction coefficient and add them up, you'll get the same result as when you would have taken the total weight and multiplied that by the friction coefficient.

So in cases like these, it's valid and easier to treat the car as a sliding block.
 
Femme_physics said:
Because it's a simple question, but in reality you do have to consider them individually?

It depends. If all the tires have the same types of surfaces, it wouldn't matter. The closer you want to get to 'reality', however, the more complications would occur. In this case, you can simply call it a block sliding on a road.

So you know your friction coefficient, you can find the acceleration and thus, 'd'. Then you can use the work-energy theorem to find how much energy is lost to friction. Then go back and determine how much time the deceleration occurred over and you'll know the power.
 
What did you do with the coefficient of friction?
I don't see it anywhere? :confused:
 
It's included in the Fs (N x 0.3)
 
Femme_physics said:
It's included in the Fs (N x 0.3)

Not consistently. I only see that you used it to calculate 6621.75 (without showing you did).

And in particular you did not use it in the formula where you find "a".
 
  • #10
wait a second, are you telling me that a = fs/m ?
 
  • #11
Femme_physics said:
wait a second, are you telling me that a = fs/m ?

Hmm, I see that you were using a = W/m = g.
That would mean...

That would mean the car was falling down, through the road!

(Yes. I'm telling you a = Fs/m. :wink:)
 
  • #12
You'll have to forgive me miss Femme... I don't know how to use TeX on this forum, yet. :)

P = W/t W is work, t is time of slide.
W = F*d F is force of friction, d is change in displacement.

_ _
[ ] <---- [ ] ->F>

F = -N*u where N is normal force, u is coefficient of friction
N = m*g m is mass of Truck, g is 9.8 m/s^2

F = Sum("All Forces") = m*a a is acceleration.
m*a = -m*g*u

a = -g*u

v' = v + a*t v' represents final velocity, v represents initial velocity

solve for t

Now we've got t.

solve for d.

d = v*t + a*t^2/2 1)

For part II

W = F*d
P = W/t 2)

WARNING: Note your velocities are in km/h... not m/s! These are the kinds of errors that will drive an ordinarily sane person to chew up their furniture.
--------------------------------------------------------------------

Here's a bonus question:

Given the heat capacities of Rubber and Concrete... what is the change in temperature of both? Do you have enough information to answer this question?

--------------------------------------------------------------------

Which language is that problem in? Is that Hebrew, or Yiddish, or something else? (Note I'm completely clueless about the history surrounding both, I've just seen them used in similar contexts)
 
  • #13
I like Serena said:
Hmm, I see that you were using a = W/m = g.
That would mean...

That would mean the car was falling down, through the road!

(Yes. I'm telling you a = Fs/m. :wink:)

So if N = 22072.5 [N]

Then

a = fs/m
a = 22072.5 x 0.3 / (22072/9.81
a = 2.943 m/s2

Now plugging this instead of the 9.81 I plugged I get 42. But in minus. Which makes sense according to the axis. Thanks :)

I'll think about how to solve B now. I'll probably use FourierFaux's reply as referrence point if I struggle. So don't go anywhere!

Till then my hat's off to you!
 
  • #14
FourierFaux said:
You'll have to forgive me miss Femme... I don't know how to use TeX on this forum, yet. :)

P = W/t W is work, t is time of slide.
W = F*d F is force of friction, d is change in displacement.

_ _
[ ] <---- [ ] ->F>

F = -N*u where N is normal force, u is coefficient of friction
N = m*g m is mass of Truck, g is 9.8 m/s^2

F = Sum("All Forces") = m*a a is acceleration.
m*a = -m*g*u

a = -g*u

v' = v + a*t v' represents final velocity, v represents initial velocity

solve for t

Now we've got t.

solve for d.

d = v*t + a*t^2/2 1)

For part II

W = F*d
P = W/t 2)

WARNING: Note your velocities are in km/h... not m/s! These are the kinds of errors that will drive an ordinarily sane person to chew up their furniture.
--------------------------------------------------------------------

Here's a bonus question:

Given the heat capacities of Rubber and Concrete... what is the change in temperature of both? Do you have enough information to answer this question?

--------------------------------------------------------------------
Thanks Fourier! Your helpful guide help me see me through to the answer :)

I got P = 163697.61

I just wonder how to I translate it to kW. I got it in kN m/s. Do I just divide by a 1000?

Which language is that problem in? Is that Hebrew, or Yiddish, or something else? (Note I'm completely clueless about the history surrounding both, I've just seen them used in similar contexts)

It's in Hebrew! Who uses Yiddish these days?!? Maybe those religious Jews in tight old-fashioned clusters

WARNING: Note your velocities are in km/h... not m/s! These are the kinds of errors that will drive an ordinarily sane person to chew up their furniture.

Always! :)

Eep at bonus question! One step at a time :D Seems to complex for me!
 
  • #15
Femme_physics said:
Now plugging this instead of the 9.81 I plugged I get 42. But in minus. Which makes sense according to the axis. Thanks :)

Right! :smile:


Femme_physics said:
I'll think about how to solve B now. I'll probably use FourierFaux's reply as referrence point if I struggle. So don't go anywhere!

Till then my hat's off to you!

Thanks! :blushing:

Have to get to work now though. See you later! :wink:
 
  • #16
It's in Hebrew! Who uses Yiddish these days?!? Maybe those religious Jews in tight old-fashioned clusters
[\QUOTE]

Very interesting... I'll have to read about the history behind that one of these days.
 
  • #17
Femme_physics said:
I got P = 163697.61

I just wonder how to I translate it to kW. I got it in kN m/s. Do I just divide by a 1000?

Eep at bonus question! One step at a time :D Seems to complex for me!


How did you get your P?


Do you have a formula table giving formulas for "work" and "power P"?
(Note that the symbol for work is usually "W", but this entirely something else than your weight "W".)


Btw, [W] = [N m/s]

and so [kW] = [kN m/s],

so there's nothing to convert. :wink:
 
  • #18
Do you have a formula table giving formulas for "work" and "power P"?
Yes this :)

FourierFaux said:
You'll have to forgive me miss Femme... I don't know how to use TeX on this forum, yet. :)

P = W/t W is work, t is time of slide.
W = F*d F is force of friction, d is change in displacement.

_ _
[ ] <---- [ ] ->F>

F = -N*u where N is normal force, u is coefficient of friction
N = m*g m is mass of Truck, g is 9.8 m/s^2

F = Sum("All Forces") = m*a a is acceleration.
m*a = -m*g*u

a = -g*u

v' = v + a*t v' represents final velocity, v represents initial velocity

solve for t

Now we've got t.

solve for d.

d = v*t + a*t^2/2 1)

For part II

W = F*d
P = W/t 2)
so there's nothing to convert.
but the answer in the manual is written as 163.7 [kW]
 
  • #19
Femme_physics said:
Yes this :)

quote

but the answer in the manual is written as 163.7 [kW]

Ah well, I'll give you the relevant formulas again, but more specifically:


Work = Force x distance

Power = Work / time


Can you apply those formulas?

That is, can you calculate the time the truck is sliding?
And the "Work" that is done by the friction force?
And finally the related "Power"?
 
  • #20
Yea I did that that's how I got the score :)

Fs x d was the force of friction I got x 42 I got

That's how I get W with units of [N x m]

And the time I got from

v' = v + a*t v' represents final velocity, v represents initial velocity

Using the "a" I got before. I think t turned out to be 1.69 [sec] or something like that.

Then I just relate those two outcomes with the power equation! Easy peasy :) Thankxxxxxxxx!
 

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