a block of mass m is placed on a rough inclined plane which makes an angle @ with the horizontal..if the incline plane is placed in an elevator moving up with a uniform vel v and the block does not move,find the work done by the frictional force in a time t....
1Work done by frictional force ...
can't we find the force on the box and the work that must have been done on it if there was no friction ( v and t are given ) and then the frictional work = work done by it if it had moved ?
mg is force downwards, and mgsin @ is the force in the dir of the plane.
The dist that must have been moved = vt
and in the dir of plane = vtsin @.
W.D = (vtsin @ ) (mgsin @) which gives the answer ...
62No varun. I am afraid not!
Then u could argue the same for pushing a wall.
if there was no friction between the wall and the ground it would have moved.. now bcos there is friction, work done will be same?!
1The work done 'on' the box = 0, i.e. the net force is 0.
The work done by 'frictional force' is mgvtsin2@
The work done by other forces acting on it is -mgvtsin2@
Therefore the net result cancels out
62No dear i dont buy this logic! :)
There is something that we are missing in this question!
Wchich box are u talking about.
62Why is the work on the block of mass zero!! it is moving up with the elevator.
1I meant relative to the elevator ... I was looking only at it as a system ( excluding the elevator ) moving up uniformly with vel. v for time t
62Dude.. sorry i made a big big big big big big big big mistake :((
i have made too many of them.. but i guess there u are right..
We have to take friciton in the upward direction and the displacemnt in the same direction..
i will post the solutin right now :)
Thanks buddy :)
BTW i still dont buy ur argument that the displacemtn because there is work done on the mass :)
62In the diagram above
we first use
N=mgcosθ
also F-mgsinθ = 0
frictional force F=mgsinθ
F.ds= f.vt=fvtsinθ= mgsinθ.vt.sinθ = mgvtsin2θ
