Solve for Distance Sliding in Socks with Friction of .250

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The discussion revolves around calculating the sliding distance on a floor with a coefficient of friction of 0.250, starting from an initial speed of 3.00 m/s. The user attempts to apply the work-energy principle, equating the kinetic energy at the start to the work done against friction. They derive the equation relating kinetic energy and frictional force, ultimately aiming to isolate the distance variable. The correct solution for the distance slid before stopping is determined to be 1.84 meters. The conversation highlights the importance of recognizing common factors in equations to simplify calculations.
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[SOLVED] Sliding in Socks

Homework Statement



Suppose that the coefficient of friction between your feet and the floor, while wearing socks, is 0.250. Knowing this, you decide to get a running start and then slide across the floor.

If your speed is 3.00 m/s when you start to slide, what distance d will you slide before stopping?

The Attempt at a Solution



We are given that coefficient of friction is .250

f = .250(n)

I really don't know to go from here.

We are given speed so, let me permit my brain to flow with the numbers :D

K= .5mv^2 may apply

.250mg = .5m(.250)^2
.250(9.8)m = .5m(.250)^2
2.45m = .003125m...and it ends there..so I just went the wrong way.

Solution is 1.84m, but I want to learn how to do this problem.
 
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ok

I got that

k1 = .5mvo^2
k2 = .5mv^2

f = umg

W = F *s

we know that f is a F so I could use that

so

W = -fd = -umgd
Wtotal = k2-k1

so I would think it would be safe to say taht

k2-k1 = -umgd

.5(mv^2) - .5(mvo^2) = -umgd

is this is all correct thusfar, how do I continue?
 
Last edited:
Looks good. Just solve for d. (Isolate d on one side.)
 
yep got it, at first, I was confused on what to do with the m, and realized that they are common factor and cancel out.
 
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