JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 6)
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 block is 0.5. The distance covered by the block before coming to rest is :
[use g = 9.8 ms$$-$$2]
4.9 m
9.8 m
12.5 m
19.6 m
Explanation
$$S = {{{u^2}} \over {2a}} = {{{u^2}} \over {2(\mu g)}}$$
$$ = {{{{(9.8)}^2}} \over {2 \times 0.5 \times (9.8)}}$$
$$ = {{9.8} \over 1}$$
$$ = 9.8$$ m
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