Block sliding Down; find kinetic coefficient

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.

None.

The Attempt at a Solution

not sure how but i keep getting: .21766

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.

None.

The Attempt at a Solution

not sure how but i keep getting: .21766

The Attempt at a Solution

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?

IDk what an FBD even is :(

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

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.

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?

Everything makes sense excpet where the W goes.

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?

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

But now what? :(

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.

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

And all three masses cancel out right?

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

And yes, all three masses cancel out.

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

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

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.

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

That looks to be correct.

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

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

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

g is always 9.8 ?
and theta is always 25.7 ?

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.

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

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.

Nope.
Dammmmmitt.