A block is pushed momentarily on a horizontal surface with initial velocity ‘v’. If ‘μ’ is the coefficient of sliding friction between the block and surface, the block will come to rest after time (‘g’ = acceleration due to gravity)

Question

A block is pushed momentarily on a horizontal surface with initial velocity ‘v’. If ‘μ’ is the coefficient of sliding friction between the block and surface, the block will come to rest after time (‘g’ = acceleration due to gravity) (A) (v/μ g) (B) (vg/μ) (C) (vμ/g) (D) (μ g/v)

Tardigrade Admin · Accepted Answer

Given, initial velocity =v Final velocity =0 Coefficient of fraction =μ For a straight line motion v-u=a t Here, 0-v=-(μ g) t [. As.μ=(a/g)] Negative sign indicates retardation ⇒ t=(V/μ g)

Q. A block is pushed momentarily on a horizontal surface with initial velocity ‘ $v$ ’. If ‘ $μ$ ’ is the coefficient of sliding friction between the block and surface, the block will come to rest after time (‘ $g$ ’ $=$ acceleration due to gravity)

$\frac{v}{μg}$

$\frac{vg}{μ}$

$\frac{vμ}{g}$

$\frac{μg}{v}$

Given, initial velocity $= v$
Final velocity $= 0$
Coefficient of fraction $= μ$
For a straight line motion
$v - u = a t$
Here, $0 - v = - (μg) t$ $[A s μ = \frac{a}{g}]$
Negative sign indicates retardation
$\Rightarrow t = \frac{V}{μg}$

Q. A block is pushed momentarily on a horizontal surface with initial velocity ‘v’. If ‘μ’ is the coefficient of sliding friction between the block and surface, the block will come to rest after time (‘g’ = acceleration due to gravity)

μgv​

μvg​

gvμ​

vμg​