Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A block $B$ is pushed momentarily along a horizontal surface with an initial velocity $v$ . If $\mu$ is the coefficient of sliding friction between $B$ and the surface, the block $B$ will come to rest after a time:

$\frac{v}{\mu g}$

$\frac{\mu g}{v}$

$\frac{g}{v}$

$\frac{v}{g}$

Step-by-Step Solution

Identify Forces: The block moves horizontally. The vertical forces are weight ( $mg$ ) downwards and normal reaction ( $N$ ) upwards. Since there is no vertical motion, $N = mg$ .
Frictional Force: The sliding (kinetic) friction force $f_k$ opposes the motion. The magnitude is given by $f_k = \mu N = \mu mg$ [NCERT Class 11, Physics Part I, Section 5.9].
Determine Retardation: According to Newton's Second Law ( $F = ma$ ), the retardation $a$ (negative acceleration) caused by friction is: $ma = -f_k = -\mu mg$ $a = -\mu g$
Kinematics: Use the first equation of motion $v_f = v_i + at$ to find the time $t$ when the block comes to rest ( $v_f = 0$ ). $0 = v + (-\mu g)t$ $\mu g t = v$ $t = \frac{v}{\mu g}$ (Reference: NCERT Class 11, Physics Part I, Chapter 5, Laws of Motion).

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