Question

A block slides with a velocity of 10 m/s on a rough horizontal
surface. It comes to rest after covering a distance of 50 m.
then the coefficient of friction between the block and surface is​

Answer 1

Explanation:

use distance formula

s=u square ÷2ukmg

Answer 2

Answer:

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

Explanation:

Given,

Initial velocity (u)       = 10 m/s

Final velocity (v)        = 0 m/s

distance moved  (s)   = 50 m

we know,

$a=-\mu g$   ...(1)

where

a is the retardation

μ is the coefficient of kinetic friction

and g is the acceleration due to gravity.

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

Retardation can be calculated by using the equation of motion,

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

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

 $a = \frac{0^{2}- 10^{2} }{2\times50} = -\frac{100}{100} = -1 \ m/s^{2}$

Now,

Using equation (1),

μ =   $\frac{a}{g}$

  $= \frac{1}{10}$

   = 0.1

ANS :

Coefficient of friction between the block and the surface = 0.1