A-block-lying-on-a-horizontal-conv-eyor-belt-moving-at-a-constant-velocity-receives-a-velocity-5m-s-at-t-0-sec-relative-to-the-ground-in-the-direction-opposite-to-the-dir-ction-of-motion-of-the-

Question Number 40396 by LXZ last updated on 21/Jul/18

A b l o c k l y i n g o n a h o r i z o n t a l c o n v -

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

v e l o c i t y r e c e i v e s a v e l o c i t y 5 m / s

a t t = 0 s e c . r e l a t i v e t o t h e g r o u n d

i n t h e d i r e c t i o n o p p o s i t e t o t h e d i r -

c t i o n o f m o t i o n o f t h e c o n v e y o r .

A f t e r t = 4 s e c, t h e v e l o c i t y o f t h e

b l o c k b e c o m e s e q u a l t o t h e v e l o c i t y

o f t h e b e l t . t h e c o e f f i c i e n t o f

f r i c t i o n b e t w e e n t h e b l o c k a n d t h e

b e l t i s 0.2 . t h e n t h e v e l o c i t y o f t h e

c o n v e y o r b e l t i s : (g = 10 m / s^{2})

(A) 13 m / s (B) - 13 m / s

(C) 3 m / s (D) 6 m / s

Commented by prakash jain last updated on 22/Jul/18

Friction force = 0.2 \times m \times g

= 2 m

acceleration (opposite to relative

motion) = - 2 {ms}^{- 2}

v v e l o c i t y i g b e l t w r t t o g r o u n d

5 + a t = - v

5 - 2 \times 4 = - v \Rightarrow v = 3