Question

a block of mass 10 kg moving with acceleration 2 metres per second square on a horizontal rough surface is shown in the figure the value of coefficient of kinetic friction is​

Answer 1

Answer:

Explanation:

Please add the figure to the question

Answer 2

Answer:

The value of coefficient of kinetic friction is 0.204.

Explanation:

As per the given data of question,

mass of block = 10 kg

acceleration = $2\frac{m}{s^{2} }$

As the figure below,

Frictional force acts in the opposite direction of the motion of force and represented by ''f''

Mathematical formula is,

$f=\mu N$

where,

$\mu$ = coefficient of friction

N = normal reaction of the body

according to figure there is a gravitational force acts on a body =mg

By balncing the forces,

N = mg

N = 10×9.8                            [ Take: $g=9.8\frac{m}{s^{2} }$ ]

N = 98$\frac{m}{s^{2} }$

and,

f = m×a

⇒ f = 10 × 2

⇒ f = 20

Now,

$\Rightarrow \mu N=20$

$\Rightarrow \mu 98=20$

$\Rightarrow \mu =\frac{20}{98}$

$\Rightarrow \mu =0.204$

Hence, the coefficient of kinetic friction is 0.204.