Velocity of wood block down an incline

[FancyNut]

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a angle. The block's initial speed is 10 m/s . The coefficient of kinetic friction of wood on wood is .200.

So that's the intro... The first question is to get the height it reachers on the incline and I got that right = 3.79 meters. During solving the first part I got delta x and it's = 1.89 meters.

it's the second part I don't know and it actually looks like the easier part.

What speed does it have when it slides back down to its starting point?

I anaylized the block from its position on top again with friction being positive (because its direction is right) and calculated the new acceleration which turned out to be -3.2 m/s^2.

Just to be sure this is it:

a_x = (u_k mgcos(30) - mgsin(30) )/m

of course I got mg out and m canceled with the bottom and the the final expression looked different..

I know the initial velocity is zero, so the final one (where it reaches the part it started on in the introduction) is the square root of 2 times acceleration times delta x... all numbers I have.

The number I got was 3.46 m/s and it turned out to be incorrect. =\


Where did I go wrong here? Thanks for any advice! :)

 

[Spectre5]

I think the second part would be more easily done using energy

Kinetic energy(initial) + potential Energy(initial) + Work = kinetic energy(final) + potential(final)

You know the initial KE, the initial PE, the work you can calculate, the final PE is 0 (if u have your datum line at the starting position), and the final KE is what you are looking for (or more correclty the velocity component of it)

 

[FancyNut]

we haven't reached those topics yet... and I don't remember anything from high school.

 

[Spectre5]

ok, where you went went was using delta x...you don't want to use delta x, you want to use the length that the box moves along, which is the hypotenuse of the triangle (or your answer from part a)

 

[FancyNut]

ok, where you went went was using delta x...you don't want to use delta x, you want to use the length that the box moves along, which is the hypotenuse of the triangle (or your answer from part a)

delta x is the hypotenuse-- I needed to get it before calculating the height it reaches for part a. 3.79 meters is the height which is the side opposite to theta = 30. I did try using 3.79 for the distance anyway and the result was still wrong. :(

 

[Spectre5]

hm...mind drawing a basic diagram? What's the answer? I think I would get 4.93 if I did the right problem

 

[Spectre5]

if the hieght is 3.79 and the hypotenuse is 1.89...then you could not have a 30 degree angle...where the sin(theta) = hieght/hypotenuse

 

[FancyNut]

Ok beware this is my own drawing of the problem... there is not graph with the question so this is just my interpetation... using my not-so-l33t microsoft paint skills...

A = 3.79 meters <--- this has to be correct because I got the answer right

delta x is the distance between the block when it was at the bottom and when it was at the top: 1.89 meters <--- also correct because I used it to get A


The block at the bottom is when it was pushed in the first part of the problem-- they wanted to know the height it reaches which is A. For the second part-- the one I'm stuck on-- they want the velocity of the block when it slides back down to its starting point. Since it slides back down velocity_initial is zero and I already know displacement so the acceleration is the last thing I need... and I calculated it 3 times already with the same result. :(

this is the shorter version of what was in the first post:

a_x = g (u_k cos(30) - sin(30) )


[EDIT]

These were my answer attempts:

4.96
3.46
5.30
4.93

 

[Spectre5]

Think about it...how can theta be 30 degrees then??

do this on your calculator:

arcsin(3.79/1.89)

that does not even exist, let alone being 30 degrees


How can the hypotenuse of a right triangle be smaller than either side?? Perhaps the question was to find the distance it moved...and if it was not, then your hypotenuse could not be 1.89...

 

[Spectre5]

Do you know the answer?

 

[FancyNut]

the 1.89 is just the delta x, the distance it moved on the incline... I was wrong for saying it's the hypotenuse.. sorry.

Do you know the answer?

no but it's very tempting to click "show answer." I'm going to do it any minute now...

 

[Spectre5]

Lol...I think u have confused me now, I have no idea what the question is anymore...if it moved by delta x and is not at a hieght of 3.79, then it must have had some initial hieght before moving?

 

[FancyNut]

ah to hell with it... it doesn't count many points in the big picture anyway...

*clicks show answer*

It's 6.97 m/s

...

 

[FancyNut]

Lol...I think u have confused me now, I have no idea what the question is anymore...if it moved by delta x and is not at a hieght of 3.79, then it must have had some initial hieght before moving?

*thinks a bit*

Yeah why I said doesn't make any sense... :(

 

[Spectre5]

lol...ok...just so u know...here it is...

the hieght is 3.79
the angle is 30degrees

so 3.79 / sin(30) = 7.58 = hypotenuse

then use that and you get 6.97 m/s

 

[FancyNut]

Ok here's the whole thing...

EDIT: didn't see your last post!

Thanks for the help man.

EDIT 2: Just noticed that screen grab is useless at that size. -_-

 

[Spectre5]

actually, I can read it :)

lol...yea...hopefully next time we won't have as much confusion.. :)

 

[FancyNut]

Just one question: I just used the kintematic equation with 7.58 and got the correct velocity... Now the 7.58 is the displacement the block had-- isn't that the SAME displacement it had when pushed UP the incline? It went up for 1.89 meters... shouldn't also come down for 1.89 meters?

Unless the 1.89 I got for part a is wrong and the solution for height was a fluke...

 

[Spectre5]

I think the solution you got for the hieght was an accident...I think you mixed a sin or cos or somethoing somewhere...the displacement is 7.58 and that is also how far it should have gone up...

 

[FancyNut]

Yeah it was a fluke... I just did it again and there is no 1.89 to be seen.

What did I get?

15.6

Twice as much as 7.8...

*sigh...*

*goes back to calculations *

 

[Spectre5]

lol...yea...that sounds a little better :)

 

[FancyNut]

OMG.

I just did get a 7.8... the reason I had it wrong because I forgot the y-component of weight is NEGATIVE!

I can rest in peace now.