Recent content by gracedescent

They go the opposite of gravity? They float you mean? Hmmmmph, yes; I think I understand where you are coming from now.

foward, yes?

Helium balloon "centrifugal" force Homework Statement Assume that you are driving down a straight road at a constant speed. A helium-filled balloon is tied to a string that is pinned to the front seat. Which way will the balloon swing when you apply the brakes? Explain why. Homework...

Homework Statement The block of ice weighs 500N. It is being pushed up an inclined plane with a height of 3m and a hypotenuse of 6m. 1a. How much force is need to push it to the top? 1b. How much work is required to push it to the top? Homework Equations W=Fd=deltaKE=deltaGPE The...

I'm sorry, I've never taken a physics class before... I wish it were as easy for me, I suppose this is simply another step! Thank you cepheid, I appreciate it. :redface:

I didn't think that this problem would pose such a problem. Is it hard or just boring? All it is, is deriving the acceleration and the kinetic coefficient on an inclinded plane with tension and frictional force... then again, I can not do it so... Ah well.

a=gsin{\theta}-\mu_k*gcos{\theta} thus, d=\frac{v_f^2-v_0^2}{2gsin{\theta}-\mu_k*gcos{\theta}} ? :D

\sum\textbf{F}=mgsin\theta-\muk*mgcos\theta Okay, I see now... Thank you.

Correct, out of the scope. I only need the accelration. Only problem, I totally agree with your remark on the normal force. But you need a mass to figure out this. In this case the skiers mass and force is omitted. I only was given the initial velocity, kinetic coefficient, and angle of descent...

When, Fx=Tcos(20)-mgsin(15)-ukn Since the uk=0.161 If I insert this into the above I obtain Fx= -1.11 So there is still something wrong with the angles I belive...

When I do how you said: Fx=Tcos(20)-mgsin(15)-ukn Fy=Tsin(20)+n-mgcos(15) ------------------------------ I obtain uk=0.141 Which is (0.020) off but something is still missing?

s= vi2/2gsin\Thetauk is that correct?

So, what your saying is that: Fx=Tcos35-mgsin15-fk=max=ma=0 Fy=Tsin35+n-mgcos15=may=0 Well I think that the tension theta of 35 and the hill theta of 15 should be separated and not subtracted by themselves?

Oh, your killing me with all these edifying intimations. Yes, indeed, I will try! :smile: Thank youu..

Well, I have that: (initial velocity) vi=20m/s (kinetic coefficient) uk=0.180 (angle) \Theta=5* So, In that: d=vi2/2ukg The only addition would have to be of sin5 at the denominator?