nice prob--must solve

ABCD is a track in the vertical plane AB and CD being smooth and concave upwards BC being an inclined plane with μs=0.3 , μk=0.2 and θ=tan-11/4. horizontal dist b/w B and C is 2 metres.
if the particle is released from rest at A find the horizontal distance between B and the point where it comes to rest.

vertical ht between B & A is 4.3 metres

A is above B and D is above A of course
dumb hint-
use energy conservation in path AB and CD

7 Answers

1
Philip Calvert ·

no one yet so there are the options

a)0
b)1/3
c)2/3
d)3/2
in metres all of them

1
greatvishal swami ·

i ll solve it tomorrow (try to)

1
greatvishal swami ·

are AB & CD both concave upward????
is mass not required?????

cant get the fig currently

1
Philip Calvert ·

no mass not reqd.
ab and cd smooth so their slope doesn.t matter jus use energy conservation
point a above b and b is the lower point of inclined plane

hotspot in the question is the part bc the inclined plane

1
skygirl ·

@philip, draw the fig naa... its not clear.. :(

1
greatvishal swami ·

ok & i was seeng da inverted image haha....

1
Philip Calvert ·

ya im trying to upload the image but its not going

see if anyone of you can do it if its clear to you

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