JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Laws of Motion

A block of mass 10 kg starts sliding on a surface with an initial velocity of

9.8 ms^{- 1}

. The coefficient of friction between the surface and bock is 0.5 . The distance covered by the block before coming to rest is :

[

use

g = 9.8 ms^{- 2}]

(A)4.9 m
(B)9.8 m✓ Correct
(C)12.5 m
(D)19.6 m

Answer: (B)

Solution

The deceleration due to friction is

a = - μg = - 0.5 \times 9.8 = - 4.9 m/ s^{2}

.

Using

v^{2} = u^{2} + 2 a s

with final velocity

v = 0

and initial velocity

u = 9.8 m/s

0 = (9.8)^{2} + 2 (- 4.9) d

d = \frac{( 9.8 ) ^{2}}{2 \times 4.9} = \frac{96.04}{9.8} = 9.8 m

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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