A-long-plank-begins-to-move-at-t-0-and-accelerates-along-a-straight-track-with-a-speed-given-by-v-2t-2-for-0-t-2-After-2-s-the-plank-continues-to-move-at-the-constant-speed-acquired-A-smal

Question Number 22456 by Tinkutara last updated on 18/Oct/17

A long plank begins to move at t = 0

and accelerates along a straight track

with a speed given by v = 2 t^{2} for 0 ⩽ t

⩽ 2 . After 2 s, the plank continues to

move at the constant speed acquired .

A small block initially at rest on the

plank begins to slip at t = 1 s and stops

sliding at t = 3 s . Find the coefficient of

static and kinetic friction between the

block and the plank .

Answered by ajfour last updated on 18/Oct/17

A c c e l e r a t i o n o f p l a n k A = 4 t

{(a c c e l e r a t i o n)}_{m a x} o f b l o c k = \frac{μ_{s} m g}{m}

= μ_{s} g

g i v e n : b l o c k b e g i n s s l i d i n g a t t = 1 s

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

e l e r a t i o n o f p l a n k e x c e e d s t h a t

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

p l a n k .

\Rightarrow μ_{s} g = 4 (1 s) \Rightarrow μ_{s} = 0.4

A t t = 1 s r e l a t i v e v e l o c i t y o f b l o c k

i s z e r o .

A t t = 2 s i t i s

= \int_{1}^{2} a_{r} d t = \int_{1}^{2} (μ_{k} g - 4 t) d t

= μ_{k} g - 6

v_{r} = u_{r} + a_{r} △ t; A f t e r t = 2 s,

a_{r} = μ_{k} g - 0

\Rightarrow v_{r} = (- 6 + μ_{k} g) + μ_{k} g △ t

s i n c e a t t = 3 s v_{r} = 0, w i t h △ t = 1 s

w e h a v e 0 = - 6 + 2 μ_{k} g

o r μ_{k} = 0.3

Commented by Tinkutara last updated on 18/Oct/17

Thank you very much Sir!