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Laws of Motion question

2007 · Q134
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Laws of Motion question

2007 · Q134

NEETPhysicsLaws of MotionMCQ+4 / −1
A block B is pushed momentarily along a horizontaly surface with an initial velocity V. If  μ\muμ  is the coefficient of sliding friction between B and the surface, block B will come to rest after a time

AIPMT 2007 Physics - Laws of Motion Question 19 English
  1. A
    gμ\muμ/V
  2. B
    g/V
  3. C
    V/g
  4. D
    V/(gμ\muμ).
View written solutionFree

Correct answer: D

Given u = V, final velocity = 0.
Using v = u + at

∴\therefore∴ 0=V−at 0 = V - at\,0=V−at

⇒\Rightarrow⇒ −a=0−Vt=−Vt - a = {{0 - V} \over t} = - {V \over t}−a=t0−V​=−tV​

f = μ\mu μR = μ\mu μmg (f is the force of friction)

∴\therefore∴ Retardation, a = μg\mu gμg

So, t=Va=Vμg t = {V \over a} = {V \over {\mu g}}t=aV​=μgV​

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