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

• Femme_physics
In summary, the car slides due to the driver activating the breaks and causing the four wheels to lock. The kinetic friction coeffecient is 0.3, and the car traveled a distance of 6621.75 meters in the sliding.
Femme_physics
Gold Member

## 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

Last edited by a moderator:
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?

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".

wait a second, are you telling me that a = fs/m ?

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. )

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)

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. )

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!

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?

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

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!

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!

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!

Have to get to work now though. See you later!

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.

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.

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]

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"?

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!

## 1. How does the distance of the sliding car affect the power needed?

The distance of the sliding car does not directly affect the power needed. Instead, it is the acceleration of the car that determines the power needed. However, a higher distance may require a longer time for the car to reach its desired speed, which can affect the overall power needed.

## 2. Can the power needed for a sliding car be calculated using a formula?

Yes, the power needed for a sliding car can be calculated using the formula P = F x v, where P is power, F is force, and v is velocity. In this case, the force can be calculated using the car's mass and acceleration, while the velocity can be determined using the distance and time taken for the car to slide.

## 3. How does the weight of the car impact the distance and power needed?

The weight of the car does not directly impact the distance and power needed. However, a heavier car may require more force and power to accelerate and maintain its speed, which can affect the distance it can slide.

## 4. Is there a maximum distance a sliding car can reach?

The maximum distance a sliding car can reach depends on various factors such as the surface it is sliding on, the initial speed, and the power needed to overcome friction. In theory, a car can slide infinitely if there is no friction or resistance, but in practical scenarios, there will always be a limit to the distance it can reach.

## 5. How does the surface affect the distance and power needed for a sliding car?

The surface can greatly impact the distance and power needed for a sliding car. A smooth surface with low friction will require less power and allow the car to slide further, while a rough surface with high friction will require more power and limit the distance the car can slide.

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