Block sliding Down; find kinetic coefficient

[cassienoelle]

Homework Statement

A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 25.7°, the block starts to slide down the incline, traveling 3.50 m down the incline in 1.50 s. Calculate the coefficient of static friction between the block and the plank.
Motion of a block:
4.81×10-1

(THAT WAS THE FIRST PART OF THE QUESTION WHICH I HAVE ALREADY FOUND.)

Part 2:
Calculate the kinetic coefficent of friction between the block and the plank.

Homework Equations

None.

The Attempt at a Solution

not sure how but i keep getting: .21766

 

[Aggression200]

Homework Statement

A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 25.7°, the block starts to slide down the incline, traveling 3.50 m down the incline in 1.50 s. Calculate the coefficient of static friction between the block and the plank.
Motion of a block:
4.81×10-1

(THAT WAS THE FIRST PART OF THE QUESTION WHICH I HAVE ALREADY FOUND.)

Part 2:
Calculate the kinetic coefficent of friction between the block and the plank.

Homework Equations

None.

The Attempt at a Solution

not sure how but i keep getting: .21766

So, first, you have a FBD correct?

I'll assume that you do.

You should have a weight force going down.
A normal force perpendicular to the plank's surface.
And a friction force going the opposite direction of motion, correct?

 

[cassienoelle]

IDk what an FBD even is :(

 

[Aggression200]

Google it real quick and you should get some good notes on it.

 

[cassienoelle]

Frozen Beverage Dispenser?!
just kidding. bhaha
Okay a Free Body Diagram. Well mmy professor told me to make one but I don't know how.
But okay, what you said makes sense. A normal force, a weight force and a friction force and then the motion.

 

[Aggression200]

Right. Now, when you draw your free body diagram, you'll notice that N is slanted in a particular direction.

You can picture the x-y axis as being slightly rotated, that way the N is now directly on the y-axis, and the f force is now directly on the x-axis.

W is now a hypotenuse going down, with an angle between the W and y-axis of 25.7 degrees.

Before I go on, does the way I (tried) to explain the FBD make sense?

 

[cassienoelle]

Everything makes sense excpet where the W goes.

 

[Aggression200]

Alright, so let me try it a different way.

Consider a normal x-y axis with N going upward directly on the y-axis and the f force going directly on the x-axis.

Now, your W force will be a diagonal line in the 4th quadrant (that is, negative y, positive x).

The angle between W and the y-axis will be 25.7 degrees. Does that make any more sense that way?

 

[cassienoelle]

That makes much more sense. :)
You're my new best friend.

 

[cassienoelle]

But now what? :(

 

[Aggression200]

Haha. Thanks, just here to try to help.

Now, you need to find the sum of the forces in the x-direction.

Well, let's start by figuring out what N will be.

Since N is pointing straight upwards, and by Newton's 3rd law, N - Wcos(theta) = 0, right?

Now, N = Wcos(theta) -> W = mg, so...

N = mg*cos(theta)

f=\muN -> \mumg*cos(theta)...\SigmaF_{x} = mg*sin(theta) - \mumg*cos(theta) = ma_{x}

Correct?

If you see an error in my reasoning, correct me.

 

[cassienoelle]

Alright, most of that makes sense. i don't have the acceleration tho to find u?

 

[cassienoelle]

And all three masses cancel out right?

 

[Aggression200]

Alright, well, that's what we need to do next.

And yes, all three masses cancel out.

 

[cassienoelle]

Okay. So now we have:
gSin(Theta) - u*g*cos(theta) = a
-->
4.25 - 8.83u = a

 

[Aggression200]

Alright, most of that makes sense. i don't have the acceleration tho to find u?

Consider that the final velocity will be the Delta x / Delta t.

So, Vf = 3.50 m / 1.50 s

Vf = 2.333 m/s


Now, use your kinematic equation...

V_{f}=V_{i}+a*t

V_{i}=0 m/s

Now, solve for a...

a = \frac{V_{f}}{t}

a = \frac{2.333}{1.50}

a = ?


After that, you can solve for uk.

 

[Aggression200]

Okay. So now we have:
gSin(Theta) - u*g*cos(theta) = a
-->
4.25 - 8.83u = a

That looks to be correct.

 

[cassienoelle]

it says I'm wrong...
i got a=1.553
so, u should equal .305
...right?

 

[Aggression200]

That should be correct. (Unless we did something wrong in the procedure).

 

[cassienoelle]

we must have done something wrong...cuz it says I'm incorrect.

 

[cassienoelle]

g is always 9.8 ?
and theta is always 25.7 ?

 

[Aggression200]

g is always 9.8 ?
and theta is always 25.7 ?

g is approximately 9.8 m/s^2

Theta will be 25.7 degrees in this problem.

uk may be .322... or .323. I'm not quite sure right now.

 

[cassienoelle]

Then i have no idea what to do. i relooked my math, and it makes sense to me.

 

[Aggression200]

Then i have no idea what to do. i relooked my math, and it makes sense to me.

Maybe you could multiply the 1.553 m/s^2 by cos(25.7). Since a would be a vector, and that would, technically, give you ax.

Try that and see if that works for you. I'm honestly confused.

 

[cassienoelle]

Nope.
Dammmmmitt.

 

[Aggression200]

I'm sorry.

I'm honestly stuck on this problem. I think I summed the forces right (since I could solve for us and get the right answer).

But, when it gets to finding the acceleration, I'm confused.

Maybe if I think about it some more...

 

[cassienoelle]

Don't be sorry. Cuz I'm juust as lost as you are. The problem isn't due till tomorrow at midnight.

 

[Aggression200]

Don't be sorry. Cuz I'm juust as lost as you are. The problem isn't due till tomorrow at midnight.

No matter how many times I do this problem, I still get .305 as uk...

So, I don't know. Maybe there was a rounding thing that was done somewhere to make it different than .305.

 

[cassienoelle]

Well thank you.
If anyone else has any help, please, I'd really appreciate it.