Question

A block sliding initially with a speed of 10 m/s on
a rough horizontal surface, comes to rest in a
distance of 7 m. What is the coefficient ofkinetic
friction between the block and the surface?​

Answer 1

Answer:

refer to the attached photo, hope you will be able to understand.

Answer 2

Answer:

A block sliding initially with a speed of 10 m/s on a rough horizontal surface comes to rest at a  distance of 7 m. What is the coefficient of kinetic  friction between the block and the surface is 0.714285

Explanation:

Given,

Initial velocity (u)       = 10 m/s

Final velocity (v)        = 0 m/s

distance moved  (s)   = 7 m

Retardation can be calculated by using the equation of motion,

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

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

 $a = \frac{0^{2}- 10^{2} }{2\times7} = -\frac{100}{14} = -7.14285 \ m/s^{2}$

we know,

$a=-\mu g$  ...(1)

where μ is the coefficient of kinetic friction and g is the acceleration due to gravity.

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

Now,

from equation (1),

μ =   $\frac{a}{g}$

   $= \frac{7.14285}{10}$

   = 0.714285

ANS :

coefficient of kinetic  friction between the block and the surface = 0.714285