Block of wood pushed up a wooden ramp

talaroue

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

A 2.37 kg wood block is launched up a wooden ramp that is inclined at a 35° angle. The block’s initial speed is 9.32 m/s. What vertical height does the block reach above its starting point?

2. Relevant equations
DYNAMICS


3. The attempt at a solution

talaroue

no matter what I do i end up with a=g(ukcos(theata)+sin(theata)) is this right how am i setting it up wrong?

Doc Al

Assuming there is friction (your first post doesn't mention any), that's the correct expression for the acceleration of the block along the incline. Was the problem statement in your first post complete?

talaroue

I believe so. I took fs=us*n....and replaced fs with us*n. Is that wrong?

talaroue

or should i assume there is no friction?

Doc Al

Your initial problem statement doesn't mention friction or give a value for μ. What makes you think there's friction involved here?

talaroue

I think friction should be involved because its wood on wood, and when wood rubs against wood there is friction. Am i just thinking to much about this problem? If there is no friction then my acceleration would be just be gsin(theata). Correct?

Doc Al

Most likely, but this is a physics problem so often things are ignored. You can only tell by context. For example, are you given a chart of coefficients of friction to use?
I suspect so.
Correct.

talaroue

We weren't given a table there is one in the book but there is no wood on wood coefficient given. Thank you Doc Al, I am going to take the appoarch without friction and see how it goes and post back on here with what I come up with thank you.

talaroue

Ok so i got the acceleration, which is negative because it slow down so can i just use Vf=Vi+(a*delta t) to find the time and then just use the kneimatics equation yf=yi+ (Viy*deltat)-(.5*g*delta t^2)? I tried that and it was wrong. Do i subsitute the gravity with the accleration i solved for?

Doc Al

Of course! Use the acceleration you figured out.

Be sure to convert your answer into the form they ask for: The vertical height, not the distance along the ramp.

You can also use a different kinematic equation that combines the two you are using into one step. Save a little work.

talaroue

well isnt that why I am using Yf and Yi in the equation that will give me how high it went?

Doc Al

No. The block moves along the ramp, you found its acceleration parallel to the ramp, and that equation will find the distance traveled along the ramp. But you can easily use that answer--with a bit of trig--to find the change in height.

talaroue

ahhhhh. Making since now!

talaroue

but wait.....wouldnt the equation i use to find the length it travels up the slope be.....
Df=Di+(Vi*delta t)-(.5*a*delta t^2). The Vi is the Intial velocity not with respect to the y axis? and then the a is the calculated a?

Doc Al

You could certainly use that if you knew (or figure out) the time. (You could also you a different kinematic formula; one that relates, a, v, and d together.)
Right. The initial velocity you were given is parallel to the ramp, and the a needed is the one you calculated.

talaroue

Ok this is all my work but it still said its wrong......