# Friction, speed & distance question

#### marshall4

How far will a person slide if they are sliding at 3m/s and the coefficient of friction between the floor and their feet is 0.5

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#### HallsofIvy

Homework Helper
There's not enough information. You would have to know the person's weight (or mass) to calculate the actual friction force (friction coefficient times weight). That would determine the distance.

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#### marshall4

Originally posted by HallsofIvy
There's not enough information. You would have to know the person's weight (or mass) to calculate the actual friction force (friction coefficient times weight). That would determine the distance.

Yes, the mass was given. 120kg

With that i found the Ff by uk*mg.

Then i used the equation F= m [(vf^2-vi^2)/2d] and rearranged it to d=[m(vf^2-vi^2)]/2F
The force that i subbed into that equation was the Ff=uk*mg, is that right???

Is there another way to do this, or did i do this wrong???

#### Doc Al

Mentor
Not wrong, just a little wasted effort. Looking at your own equations, you can see that the mass cancels (as chroot pointed out) so it really is extraneous information.

Resist the temptation to start plugging in numbers prematurely. Figure things out symbolically as much as you can, then plug in the numbers.

#### PrudensOptimus

I am not sure about my approach, but here is what i think:

&Delta;x = ( v^2 - v(initial)^2 ) / 2a

&Sigma;Fx = max = Ff

max = &mu;kFN * cos 180deg

since cos 180deg = -1, Ff is a negative quantity.
and the mass m cancels out.

ax = - 0.5 * 9.8 = -5 m/s

plug back in &Delta;x equation:

&Delta;x = abs(-9/10) = 9/10 m.

The sliding guy will travel 0.9 meters.

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#### marshall4

Originally posted by Doc Al
Not wrong, just a little wasted effort. Looking at your own equations, you can see that the mass cancels (as chroot pointed out) so it really is extraneous information.

Resist the temptation to start plugging in numbers prematurely. Figure things out symbolically as much as you can, then plug in the numbers.

What does the mass cancel out with?

#### chroot

Staff Emeritus
Gold Member
F = [mu] N = [mu] m g

F = m a

[mu] m g = m a

a = - [mu] g (negative sign because the slider is slowing down)

v(t) = v0 + a t

v(t) = v0 - [mu] g

When he stops, his velocity is zero:

0 = v0 - [mu] g

t = v0 / ([mu] g)

s(t) = v0 t + 1/2 a t2

s(t) = v0 (v0 / ([mu] g) ) - 1/2 [mu] g * (v02 / ([mu] g)2 )

s(t) = 1/2 v02 / ([mu] g)

Does this help?

- Warren

#### PrudensOptimus

Originally posted by marshall4
What does the mass cancel out with?
mass cancels out with mass. it's a mathematical "trick".

#### Doc Al

Mentor
Originally posted by marshall4
What does the mass cancel out with?
With itself, of course! (D'oh!)

Using conservation of energy:

(all the KE goes into work against friction)

Fd=1/2m v^2
F=umg

umgd = 1/2mv^2 (the mass cancels here)

d= v^2/(2ug)

#### toa

The answers posted seem to be in nice accordance with the physics I was taught in high school. However, in practice they fail dismally, unless one at least brings the area of contact between the two surfaces into the equation. There is a reason formula one cars have tyres as wide as a barn door, and that the trimming of cars usually involves changing to wider tyres. I think there is some relationship which dictates that the greater the area, the greater the friction becomes. Does anyone know anything about this?

#### chroot

Staff Emeritus
Gold Member
Originally posted by toa
The answers posted seem to be in nice accordance with the physics I was taught in high school. However, in practice they fail dismally, unless one at least brings the area of contact between the two surfaces into the equation. There is a reason formula one cars have tyres as wide as a barn door, and that the trimming of cars usually involves changing to wider tyres. I think there is some relationship which dictates that the greater the area, the greater the friction becomes. Does anyone know anything about this?
Tires are quite special, because the deform, heat up, and become greasy as they are used. That does not mean that the area is important in determining friction -- in general, it is not.

- Warren

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