Given initial velocity u=9.8ms−1.
Acceleration a=μg=0.5×9.8=4.9ms−2
Using, v2=u2+2as, here, final velocity v=0.
So, 0=(9.8)2+2(4.9)s⇒s=9.8m
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 block is 0.5. The distance covered by the block before coming to rest is :[use g=9.8ms−2]
Held on 24 Jun 2022 · Verified 6 Jul 2026.
9.8m
4.9m
12.5m
19.6m
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