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I'm trying to work a physics problem about how to find the coefficient of static friction between two objects when given the initial velocity and distance. Specifically, the problem I am working is as follows:

A crate is carried in a pickup truck traveling horizontally at 14.2 m/s. The truck applies the brakes for a distance of 25.8 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed if the crate does not slide?

I do know that the formula that has to be used here is $F_F=µF_n$, but I am unsure as to how it should be applied. I really just need somewhat of a hint to get to where I can plug numbers into that formula, but I am unsure as to how I can get these numbers given the context of the problem. Thank you!

Edit: I'm not really needing the answer to this specific problem, but rather, how I can solve a question like the one above is all I really need. I'm just confused when given initial velocity and distance.

J.W.F.
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  • You know $|F_n| = g$. – copper.hat Oct 21 '13 at 17:24
  • @copper.hat: So, that means that you already have some of formula, being $F_F=µ(9.8)$. And would the force of friction be greater than the initial velocity because you're de-accelerating? – J.W.F. Oct 21 '13 at 17:28
  • Well, you need to do some work, not just look for a formula. Given uniform acceleration (deceleration in this instance), you can figure out a formula relating initial, final speed, distance and acceleration. This gives the acceleration. Then the forward acceleration is related to force by mass and similarly the normal force is related to $g$ by mass. This will give $\mu_\text{static}$. – copper.hat Oct 21 '13 at 17:32

2 Answers2

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Since you have the physics tag, here is another approach:

The work done is $F_F d$, where $d$ is the distance travelled. The initial energy is $\frac{1}{2} m V_0^2$ (potential energy is constant here), where $m$ is the truck mass and $V_0$ the initial speed. The final energy is zero. We have $F_F = \mu F_N = \mu m g$, where $g$ is the acceleration due to gravity. This gives $\mu m g d = \frac{1}{2} m V_0^2$, or $\mu = \frac{1}{2} \frac{v_0^2}{dg}$.

Using the numbers you provided, $\mu = \frac{1}{2} \frac{(14.2)^2}{(25.8)(9.81)} \approx 0.4$.

copper.hat
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You need to compare the force due to the truck's deceleration and the force due to friction. You know the force due to friction is (at most) $\mu m g$, so let's find the force due to deceleration.

You know, from kinematics, that

$$ v_f = v_0 + at_f $$ and $$ x_f = x_0 + v_0 t_f + \dfrac{1}{2}at_f^2 $$ where $x_0,v_0$ indicate initial position/velocity, and $x_f,v_f,t_f$ final position, velocity, time.

We want to know what acceleration the truck undergoes, and we know $v_0 = 14.2$, $v_f = 0$, $x_f = 25.8$ (if we take $x_0 = 0$).

So, in fact, we have

$$ 0 = 14.2 + at_f $$

$$ 25.8 = 14.2t_f + \frac{1}{2}at_f^2. $$

You can solve these equations for $t_f$ and $a$ ($a$ is what you're interested in), and the force from deceleration is $ma$.

(Note: there's also the formula $v_f^2 = v_0^2 + 2a(x_f - x_0)$, and you could simply start here to make the problem easier. The work we did above could be used to derive this formula).

BaronVT
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