Find the kinetic friction coefficient

[beborche]

Homework Statement

A pig is sliding down a ramp that has an angle of 35 degrees from the horizontal plane. If the friction coefficient was 0 the ride would only take half the time. What is the kinetic friction coefficient between pig and ramp?

Homework Equations

F_net = ma
v² = u² + 2as
a = \Deltav/t
f_k = u_kN = u_kmgsin\theta

The Attempt at a Solution

Since I wasn't given a length of the ramp, I decide it to be s=10 meters (the length shouldn't matter, I just use the value so I can perform calculations).
u = 0 m/s
s = 10 m
a = gsin\theta = 5.62 m/s²
v = \sqrt{2as} = \sqrt{2*5.62*10} = 10.6 m/s
t = (v-u)/a = 10.6/5.62 = 1.87 s
This is the time it takes without any friction. Now let's double this time, and find the acceleration. Then, through the expressions of the forces, solve for u_k:

t = 3.74 s
u = 0 m/s
s = 10 m
v = s/t = 2.67 m/s
a = (v-u)/t = 0.71 m/s²

F = ma = mgsin\theta - u_kgcos\theta
Rearrange and you'll get this
u_k = \frac{a + gsin\theta}{gcos\theta} = \frac{0.71 + 9.8sin(35)}{9.8cos(35)} = 0.787

This answer isn't correct, but it's rather close (probably by coincidence). Anyhow, can anyone hint me in the right direction? Where am I going wrong?
Thanks

 

[gneill]

First, you should use more decimal places in your intermediate results. Maybe three or four. This will guard against rounding errors creeping into 'sensitive' calculations. Also, you might want to use g = 9.807 m/s² for gravity.

The formula that you arrived at for μ_k is not quite right; it produces a negative value. This is because the numerator should be negated.

 

[beborche]

Ah you're right. I see the mistake.

F = ma = mgsin\theta - u_kgcos\theta
Rearrange and you'll get this
u_k = \frac{a + gsin\theta}{gcos\theta} = \frac{0.71 + 9.8sin(35)}{9.8cos(35)} = 0.787

should be

F = ma = mgsin\theta - u_kmgcos\theta
Rearrange and you'll get this
u_k = \frac{a - gsin\theta}{-gcos\theta} = \frac{0.71 - 9.8sin(35)}{-9.8cos(35)} = 0.613

When I got 0.613 as my final answer I used 5+ decimals and I used 9.807 for g. This is closer to the correct answer, but I'm quite sure I won't pass with this answer.

The correct answer is 0.52
Any ideas?

 

[gneill]

I would change the approach to finding the acceleration for the friction case. Using an average velocity to find acceleration is not good.

Instead, go directly from the kinematic formula s = (1/2)at². You're given t and s, so find a.

 

[beborche]

Thank you, I solved it!
Not by finding acceleration in the case of friction first, but thanks to you reminding me of that what I calculated was the average velocity. These two lines messed it up:
v = s/t = 2.67 m/s
a = (v-u)/t = 0.71 m/s²
Where it says (v-u), it should say 2*v. Since v = average velocity and u = 0, then 2*v is the change of velocity. And right we are! 0.52 is the answer.

Again thanks for your help! :-]