A-plank-of-mass-M-kg-is-sliding-on-the-smooth-horizontal-surface-with-constant-velocity-of-10-ms-1-A-another-block-of-mass-M-kg-is-gently-placed-on-it-The-coefficient-of-friction-between-the-blo

Question Number 24168 by Tinkutara last updated on 13/Nov/17

A plank of mass M kg is sliding on the

smooth horizontal surface with constant

velocity of 10 {ms}^{- 1} . A another block of

mass M kg is gently placed on it . The

coefficient of friction between the

block and the upper surface of the plank

is 0.2 . Assuming that plank is long

enough such that the block does not fall

from it . The velocity - time graph of the

block is [Take g = 10 m / s^{2}]

Commented by ajfour last updated on 14/Nov/17

Commented by Tinkutara last updated on 14/Nov/17

Why it only increases upto 2.5 s ?

Commented by ajfour last updated on 14/Nov/17

f r o m c o n s e r v a t i o n o f m o m e n t u m

f o r (p l a n k + b l o c k) : w e h a v e

M u = 2 {M v}_{f i n a l c o m m o n v e l o c i t y}

\Rightarrow v_{f i n a l c o m m o n v e l o c i t y} = \frac{u}{2} = 5 m / s

t i m e t a k e n f o r b l o c k t o r e a c h t h i s

v e l o c i t y i s o b t a i n a b l e f r o m :

v = 0 + μ g t

5 = (0.2 \times 10) t

t = 2.5 s .

A f t e r t h i s t h e y m o v e t o g e t h e r

w i t h t h e s a m e v e l o c i t y a n d t h e

m u t u a l f r i c t i o n n o m o r e a c t s .

Commented by Tinkutara last updated on 14/Nov/17

Thank you very much Sir!