# Determining kinetic friction coefficient

So here's my problem:

An object of mass m = 4 kg starts from rest from the top of a rough inclined plane a with θ = 30◦ incline, L = 20 m, and an unknown coefficient of friction, μk. The speed of the object at the bottom of the inclined plane is v = 10 m/s.

A) Find the work done by friction What I did is I know $W_{non conserved}=E_{F}-E_{i}$, or kinetic energy minus potential energy. I plugged in 0.5(4)(100)-4(9.8)(10) to get -192 J

B) Find the coefficient of friction. I know that $W= |F||d|cos\theta$, but since the ball is rolling down we have to add a negative sign before the force, F. So I have -|F|10cos(30)=-192 J, which is what I got from the above section. I find that force is 22.17 J and this should equal mass x acceleration in the x direction. The mass is 4 kg which means the acceleration must be at $5.54 m/s^2$.

I know from drawing a free body diagram that $\mu_{k}g cos(30)+g sin(30)=a_x$. If i plug in the acceleration 5.54 m/s^2 from above and solve for the coefficient I get 0.076.

Is this approach using the principles correctly and does it make sense?

• My knowledge of physics is thin. You might have better luck at physics.stackexchange.com Nov 15, 2017 at 17:22
• part a is Ok. in b why d=10.? Nov 15, 2017 at 17:48
• Good catch; d should be 20, since the mass travelled 20 m. Nov 16, 2017 at 4:59

## 1 Answer

My physics is pretty rusty, but this is what I remember

Solving for the frictional force:

$W = F\cdot d\\ F = -9.6\\ \mu (mg)\cos 30 = |F|\\ \mu = 0.28$

Solving for the impact of acceleration.

$F = ma\\ a = -2.4$

Net acceleration $= 9.8 \sin 30 - 2.4 = 2.5 \frac {m}{s^2}$

And accelerating over this distance.

$20 = \frac 12 (2.5) t^2$

$t = 4$ seconds

$v = at = 10 \frac {m}{s}$