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marshall4

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- Thread starter marshall4
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marshall4

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HallsofIvy

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chroot

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Sorry, Halls, that's wrong. The mass cancels.Originally posted by HallsofIvy

See this thread:

https://www.physicsforums.com/showthread.php?s=&threadid=8395&perpage=12&pagenumber=1

- Warren

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marshall4

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Originally posted by HallsofIvy

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?

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Doc Al

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

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PrudensOptimus

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

Δx = ( v^2 - v(initial)^2 ) / 2a

ΣF_{x} = ma_{x} = F_{f}

ma_{x} = μ_{k}F_{N} * cos 180deg

since cos 180deg = -1, F_{f} is a negative quantity.

and the mass m cancels out.

a_{x} = - 0.5 * 9.8 = -5 m/s

plug back in Δx equation:

Δx = abs(-9/10) = 9/10 m.

The sliding guy will travel 0.9 meters.

Δx = ( v^2 - v(initial)^2 ) / 2a

ΣF

ma

since cos 180deg = -1, F

and the mass m cancels out.

a

plug back in Δx equation:

Δx = abs(-9/10) = 9/10 m.

The sliding guy will travel 0.9 meters.

Last edited:

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marshall4

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Originally posted by Doc Al

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

What does the mass cancel out with?

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chroot

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F = m a

[mu] m g = m a

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

v(t) = v

v(t) = v

When he stops, his velocity is zero:

0 = v

t = v

s(t) = v

s(t) = v

s(t) = 1/2 v

Does this help?

- Warren

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PrudensOptimus

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Originally posted by marshall4

What does the mass cancel out with?

mass cancels out with mass. it's a mathematical "trick".

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Doc Al

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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)

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toa

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chroot

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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.Originally posted by toa

- Warren

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