@lemmaThe question states that the coefficient of friction is tan(A). I believe this is the mu(kinetic). Why assume that it is the coefficient of static friction?

QuoteOriginally posted by: lemma@vixennow i get the point....wen the body begins moving...its kinetic friction everywhere....so the m*g*sin(a)>mu(kinetic)*m*g*cos(a)....agreed...this will trigger the motion along downward direction....and this unbalanced force along downward ll continue till the motion stops...though the vz=0 ll happen at sometime.....vx will never become zero because the body will be accelerated downwards with acc=g*sin(a)-mu(kinetic)*g*cos(a)....and acc ll always be positive and hence by this proposition velocity will be infinite after a very long time....P.S: mu(kinetic) < mulemma, I'm sorry, but I think you still didn't get the point. First of all, what do you call the "coefficient of static friction"? Whatever it is, where does the word "static" come from in this problem? The body is never at rest!

@vixen,shryuik,othersokies....agreed....i was always thinking of it as coeff of static friction....thanks for pointing out that it is mu(kinetic)...but if it mu(kinetic) = tan(a)....mu(kinetic)*m*g*cos(a) = m*g*sin(a)....can u please explain how there will be an unbalanced force in direction X?until the point mass is imparted some velocity in the z direction...it is at rest due to static friction.....the moment it is set in motion kinetic friction comes into play but this force mu*m*g*cos(a)=m*g*sin(a)...so no unbalanced force...i think someone has mentioned that the force of friction is always opposite to direction of motion..agreed..this is wat opposes vz..but u must also realise that as long as the body is in the inclined plane it has a tendency to slide down due to m*g*sin(a).....and the force of friction ll not only come into play when the object is moving but also when the object has a tendency to move...so friction ll act both along z-direction and x-direction...so it will not move in the x-direction

QuoteOriginally posted by: vixenWe can easily choose units such that the acceleration due to gravity along the plane = acceleration due to friction = 1. The DE is now-cos(th) = d/dt ( v cos(th) )1 - sin(th) = d/dt ( v sin(th) )where th is the angle of the velocity to the horizontal. Eliminating dt, integrating and taking the limit as th tends to pi/2 gives mefinal v = V0/2.I totally agree! Great solution. I spotted the correct co-ordinates, but failed to see the need to find v(\theta) rather than \theta and v as functions of time until reading your post.

QuoteOriginally posted by: lemma@vixen,shryuik,othersokies....agreed....i was always thinking of it as coeff of static friction....thanks for pointing out that it is mu(kinetic)...but if it mu(kinetic) = tan(a)....mu(kinetic)*m*g*cos(a) = m*g*sin(a)....can u please explain how there will be an unbalanced force in direction X?until the point mass is imparted some velocity in the z direction...it is at rest due to static friction.....the moment it is set in motion kinetic friction comes into play but this force mu*m*g*cos(a)=m*g*sin(a)...so no unbalanced force...i think someone has mentioned that the force of friction is always opposite to direction of motion..agreed..this is wat opposes vz..but u must also realise that as long as the body is in the inclined plane it has a tendency to slide down due to m*g*sin(a).....and the force of friction ll not only come into play when the object is moving but also when the object has a tendency to move...so friction ll act both along z-direction and x-direction...so it will not move in the x-directionlemma, we are physicists, aren't we? setup the experiment and prove or throw away your or our conclusions!

haha....its been almost 7 yrs since i last looked at a mechanics related problem....it was a really nice experience...thanks ppl....