The-force-acting-on-the-block-is-given-by-F-5-2t-The-frictional-force-acting-on-the-block-at-t-2-s-The-block-is-at-rest-at-t-0-

Question Number 20501 by Tinkutara last updated on 27/Aug/17

The force acting on the block is given by

F = 5 - 2 t . The frictional force acting

on the block at t = 2 s . (The block is at

rest at t = 0)

Commented by Tinkutara last updated on 27/Aug/17

Commented by ajfour last updated on 27/Aug/17

J = \int_{0}^{t} [(5 - 2 t) - f] d t

J = 5 t - t^{2} - 2 t = 3 t - t^{2}

J = 0 w h e n t = 0 a n d t = 3 s

t h e r e f o r e b l o c k i s m o v i n g u n t i l

t = 3 s; f r i c t i o n a t t = 3 s,

i s e q u a l t o f = μ m g = 0.2 \times 1 \times 10

= 2 N .

Commented by Tinkutara last updated on 27/Aug/17

Thank you very much Sir!

Commented by Tinkutara last updated on 27/Aug/17

And another thing can you explain

that maximum frictional force is μ m g

= 2 N and at t = 0, applied force = 5 N,

so how the block remains at rest ? How

2 N stops 5 N ?

Commented by ajfour last updated on 27/Aug/17

a t t = 0 b l o c k w a s a t r e s t, (g i v e n)

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

f r i c t i o n t r i e s t o r e s i s t, a n d e v e n

e x t e r n a l f o r c e d e c r e a s e s a s t i m e

p r o c e e d s . t h e b l o c k g a i n s s p e e d

u n t i l F = 5 - 2 t = μ m g = 2

t h a t i s t i l l t = 1.5 s . A f t e r t h i s

f r i c t i o n i s m o r e = 2, (\leftarrow); b l o c k

s l o w s d o w n, f o r c e k e e p s d e c r e a s i n g

a n d e q u a l s z e r o a t t = 2.5 s

a n d t h e r e a f t e r n e g a t i v e,

b l o c k s t o p s a t t = 3 s

a g a i n, r e s t s f o r a l i t t l e t i m e

(m e a n w h i l e f o r c e c h a n g e s a n d

f r i c t i o n r e v e r s e s d i r e c t i o n,

b l o c k t h e n m o v e s l e f t,

f r i c t i o n a c t s r i g h t w a r d s . .

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