Question

A block pushed horizontally with a velocity of 8 m/s and then comes at rest after sliding a horizontal distance of 16 m. What is the coefficient of kinetic friction between the block and the ground?

Answer 1

Explanation:

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Answer 2

Answer:

A block is pushed horizontally with a velocity of 8 m/s and then comes at rest after sliding a horizontal distance of 16 m. The coefficient of kinetic friction between the block and the ground is 0.2

Explanation:

Given,

Initial velocity (u)       = 8 m/s

Final velocity (v)        = 0 m/s

distance moved  (s)   = 16 m

we have the equation,

$a=-\mu g$   ...(1)

where

a - The retardation

μ - Coefficient of kinetic friction

and g  -  Acceleration due to gravity.

g = 10 $\ m/s^{2}$

Retardation can be calculated by using the equation of motion,

$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aS}$

$a = \frac{v^{2}- u^{2} }{2s}$

 $a = \frac{0^{2}- 8^{2} }{2\times16} = -\frac{64}{32} = -2 \ m/s^{2}$

Now,

Using equation (1),

μ =   $\frac{a}{g}= \frac{2}{10}$

   = 0.2

ANS :

Coefficient of friction between the block and the surface = 0.2