Question

A block slides with an velocity of 10m/s on a rough horizontal surface. It is brought to rest after covering a distance of 50m. Then coefficient of friction is (g = 10m/s)​

Answer 1

Answer:

Given,

Initial velocity, u=10ms

−1

Displacement, s=50m

Final velocity, v=0

Acceleration of friction, a=μg=10μms

−2

Apply second kinematic equation of motion,

v

2

−u

2

=2as

0−10

2

=2(−10μ)×50

μ=0.1

Hence, coefficient of friction is 0.1

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Answer 2

Answer:

A block slides with a velocity of 10m/s on a rough horizontal surface. It is brought to rest after covering a distance of 50m. Then coefficient of friction is $0.1$

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 body $v=10m/s$

the stopping distance $d=50m$

substitute the given values to get the coefficient of friction

μ$=\frac{v^{2} }{2*d*g}=\frac{10^{2} }{2*50*10}=0.1$