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I am not sure how to approach a problem involving the friction coefficient when only given initial velocity, final velocity, and distance traveled. I could combine the friction and kinematic equations, but I am only familiar with doing that for one velocity...

Here is the specific problem I had trouble with:

A cardboard box of unknown mass is sliding upon a mythical friction-less surface.

The box has a velocity of 4.56 m/s when it encounters a bit of friction. After sliding 0.700m, the box has a velocity of 3.33 m/s.

What is the coefficient of friction of the surface?

How do I solve this problem in particular, and what is the general procedure to solve problems like this?

amWhy
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tyger2020
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3 Answers3

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You write an equation that incorporates the information in the problem. Here you are supposed to assume the friction is a constant negative acceleration. If you let $x=0, t=0$ be where it encounters the friction, it is $x=v_0t-\frac 12 at^2, v=v_0-at$

The friction coefficient is the friction force divided by the weight. We are not given the weight here. How do you overcome that?

Ross Millikan
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  • Frictional force: F = ma = mg $\mu$. $\mu$ = a/g.Makes sense? – Peter Szilas Aug 23 '17 at 22:49
  • @PeterSzilas : yes, that is right – Ross Millikan Aug 23 '17 at 22:50
  • @RossMillikan if friction force = a/g, how might I find acceleration if I do not have change in time? With one equation, I got -6.93 for acceleration. Is this correct? Because I did -6.93 / 9.8, got -0.707, and was marked incorrect... – tyger2020 Aug 23 '17 at 23:59
  • I gave you two equations in two unknowns, $a$ and $t$. You should solve them simultaneously with the two velocities you are given. You have $v_0=4.56, v(t)=3.33$ when $x=0.7$ – Ross Millikan Aug 24 '17 at 02:51
  • @RossMillikan Ok, so now I got acceleration = 6.932 and time = 0.19, but how do I use that information to find the friction coefficient? I want to know normal force to use in my equation, but I don't have mass... – tyger2020 Aug 24 '17 at 20:02
  • Peter Szilas' comment is spot on. The mass divides out when you look at the friction coefficient. – Ross Millikan Aug 24 '17 at 22:39
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Friction is (at a first approximation) a pressure tangential to the surface of contact between two bodies, directed against the movement, and proportional to the contact pressure, by the coefficient of friction $\mu$.
Thus totally it is equal to $\mu mg$, and will exert a deceleration equal to $\mu g$. The law of accelerated motion gives $v_1=v_0-\mu g \Delta t$, i.e. $\Delta t=(v_0-v_1)/(\mu g)$.
Then $s=(v_0-1/2\mu g \Delta t)\Delta t$.
Can you combine the two and get $\mu g$ and thus $\mu$ ?

G Cab
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There are a couple of different ways to approach this. One way is to consider the rate of change of the box's speed caused by friction. But given the setup of this problem, I prefer a "work = change in energy" approach instead.

For the "work" approach, compute the work done by the box against the force of friction over the given distance. That amount of work is an amount of energy expended by the box, which comes out of the box's kinetic energy. That is, $$ \text{kinetic energy of box before} = \text{work performed} + \text{kinetic energy of box afterwards.} $$

Now you just have to figure out formulas for all three of those things. The formulas involve some known quantities (distance, speed before, speed after, gravity) and some unknown quantities (mass, coefficient of friction), but I usually like to set up the equations using variables for everything (as if all these things were unknown) and then write the equations again with the known things filled in.

Now you have to hope that after setting up the equation, you can find some kind of cancellation that lets you eliminate the mass. Hint: you can.

David K
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