Question

43. A block of mass 10 kg, moving with acceleration 2 m/s2 on horizontal rough surface is shown in figure
→ a= 2 m/s
> F= 40 N
The value of coefficient of kinetic friction is
(1) 0.2
(3) 0.5
(2) 0.4
(4) 0.1​

Answer 1

Answer:

(i)0.2=coefficient of kinetic friction

Explanation:

formula: [f(k)= μk N]

so, we first have to find the value of N and fk to solve the problem:--

N=mg=10*10=100    [taking g=10m/s²]-------------(i)

now, force=mass*a

F-fk=ma

40-fk= 10*2

40-fk=20

-fk=20-40

-fk=-20

fk=20  ---------------(ii)

substitute value of (i) and (ii) in the formula:

we get,

μk=fk/N

    = 20/100

    = 0.2

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Answer 2

Therefore the value of the coefficient of kinetic friction is 0.2. ( Option-1 )

Given:

Mass of the block = 10 kg

Acceleration of the block = 2 m/s²

Force applied = 40 N

To Find:

The value of the coefficient of kinetic friction.

Solution:

The given question can be solved very easily as shown below.

Given that,

Mass of the block = m = 10 kg

Acceleration of the block = a = 2 m/s²

Force applied = F = 40 N

As the block is moving on a  horizontal floor, D'Alembert's Principle can be applied.

By applying D'Alembert's Principle,

⇒ F - μN = ma

Where μ = Coefficient of kinetic friction

N = Normal reaction on the block = weight of the block = mg

⇒ F - μmg = ma

On substituting values in the above equation, we get,

⇒ 40 - ( μ × 10 × 9.81 ) = 10 × 2

⇒ 40 - ( 98.1μ ) = 20

⇒ 40 - 20 = 98.1μ

⇒ μ = 20 / 98.1 = 0.2

Therefore the value of the coefficient of kinetic friction is 0.2.

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