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A block slide down an inclined plane of slope angle theta with constant velocity. it is the projected up the same plane with an initial velocity u. How far up the incline wil it move before coming ti rest.??
Dear Alpha see the pic for solution All the best. AKASH GOYAL AskiitiansExpert-IITD Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Dear Alpha
see the pic for solution
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
let the angle of incline is @.. it is given that block is moving with constant velocity when slides down the incline so net force along the incline is zero or friction force is equal to force due to gravity along incline.. or friction force = mgsin@ ......1 when particle moves upward both friction and gravity acts in opposite direction so net force = mgsin@ + f(friction force) =mgsin@+mgsin@=2mgsin@ (from eq 1) net force = m(net retardation) a(retardation) = -2gsin@ ..............2 (-ve sign is coz force acts in opp direction of motion) applying , v2 = u2 + 2as finally particle stops so v becomes 0.. so 0 = u2 -2(2gsin@)s s= u2/4gsin@ .......... ..3 so the block moves this much distance before coming to rest .....
let the angle of incline is @..
it is given that block is moving with constant velocity when slides down the incline
so net force along the incline is zero or friction force is equal to force due to gravity
along incline..
or friction force = mgsin@ ......1
when particle moves upward both friction and gravity acts in opposite direction so
net force = mgsin@ + f(friction force)
=mgsin@+mgsin@=2mgsin@ (from eq 1)
net force = m(net retardation)
a(retardation) = -2gsin@ ..............2 (-ve sign is coz force acts in opp direction of motion)
applying , v2 = u2 + 2as
finally particle stops so v becomes 0..
so 0 = u2 -2(2gsin@)s
s= u2/4gsin@ .......... ..3
so the block moves this much distance before coming to rest .....
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