Question

A block sliding on rough horizontal plane with velocity of 5m/s come for the rest 12.5m. the coefficient of friction between the surface and the block is (g=10m/s) in brainly.in

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

A block sliding on a rough horizontal plane with a velocity of 5m/s come for the rest of 12.5m. the coefficient of friction between the surface and the block is (g=10m/s) $0.1$

Explanation:

When a body moves along a rough surface it travels a distance before it comes to rest, this distance is called stopping distance

for a body moving with velocity $v$ through a rough surface with a coefficient of friction μ the stopping distance is given by the formula

$d=\frac{v^2}{2\mu g}$

where the acceleration due to gravity $g=9.8m/s^{2}$≈$10m/s^{2}$

from the question, it is given that

the velocity of the sliding block $v=5m/s$

the stopping distance $d=12.5m$

substitute the given values to get the coefficient of friction

μ$=\frac{v^{2} }{2*d*g}=\frac{25}{2*12.5*10}=0.1$