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 ]

Question

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 (C) 12.5 m (D) 19.6 m

Tardigrade Admin · Accepted Answer

a =-μ g =-0.5 × 9.8=-4.9 m / s 2 d =( v 2/2 a )=(9.8 × 9.8/2(4.9)) =9.8 m

Q. A block of mass $10 k g$ starts sliding on a surface with an initial velocity of $9.8 m s^{- 1}$ . The coefficient of friction between the surface and bock is $0.5$ . The distance covered by the block before coming to rest is : [use $g = 9.8 m s^{- 2}$ ]

$4.9 m$

$9.8 m$

$12.5 m$

$19.6 m$

$a = - μg = - 0.5 \times 9.8 = - 4.9 m / s^{2}$
$d = \frac{v ^{2}}{2 a} = \frac{9.8 \times 9.8}{2 ( 4.9 )}$
$= 9.8 m$

Q. A block of mass 10kg starts sliding on a surface with an initial velocity of 9.8ms−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=9.8ms−2 ]

4.9m

9.8m

12.5m

19.6m