# Block sliding Down; find kinetic coefficient

1. Sep 20, 2011

### cassienoelle

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
None.

3. The attempt at a solution
not sure how but i keep getting: .21766
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 20, 2011

### Aggression200

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?

3. Sep 20, 2011

### cassienoelle

IDk what an FBD even is :(

4. Sep 20, 2011

### Aggression200

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

5. Sep 20, 2011

### 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.

6. Sep 20, 2011

### 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?

7. Sep 20, 2011

### cassienoelle

Everything makes sense excpet where the W goes.

8. Sep 20, 2011

### 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?

9. Sep 20, 2011

### cassienoelle

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

10. Sep 20, 2011

### cassienoelle

But now what? :(

11. Sep 20, 2011

### 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=$\mu$N -> $\mu$mg*cos(theta)....

$\Sigma$F$_{x}$ = mg*sin(theta) - $\mu$mg*cos(theta) = ma$_{x}$

Correct?

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

12. Sep 20, 2011

### cassienoelle

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

13. Sep 20, 2011

### cassienoelle

And all three masses cancel out right?

14. Sep 20, 2011

### Aggression200

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

And yes, all three masses cancel out.

15. Sep 20, 2011

### cassienoelle

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

16. Sep 20, 2011

### Aggression200

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

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.

17. Sep 20, 2011

### Aggression200

That looks to be correct.

18. Sep 20, 2011

### cassienoelle

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

19. Sep 20, 2011

### Aggression200

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

20. Sep 20, 2011

### cassienoelle

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