Question

With 9N force body of mass10kg moves on horizontal surface then find coefficient of kinetic friction

Answer 1

$\boxed{{\blue{Answer}}}$

$\nu_k = 0.09$

$\boxed{{\blue{Explanation}}}$

Given:

• F = 9N
• m = 10kg

To Find:

Co-efficient of Frictional Force

Solution:

W.k.t,

$\boxed{{\red{F = ma}}}$

$9 = 10a$

$a = \dfrac{9}{10}$

$a = 0.9 m{s}^{-2}$

Now,

$\boxed{{\red{F= \nu N}}}$

$ma = \nu_k mg$

$a = \nu_k g$

$0.9 = \nu_k (10)$

$\nu_k = \dfrac{0.9}{10}$

$\nu_k = 0.09$

$\texttt{\blue{Aravind}\:\red{Reddy}....!}$