Question

NEET &
mass 10 kg, moving with acceleration 2 m/s2 on he
acceleration 2 m/s on horizontal rough surface is shown in figure
Laws of Motion
58. A block of mass 10
→ a= 2 m/s?
> F= 40 N
of coefficient of kinetic friction is
The value of coefficier
20:27
(1) 0.2
(3) 0.5
(2) 0.4
(4) 0.1
-
h
a
ffaiantar 1-​

Answer 1

Answer:

(1)0.2 is the 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

Hope It Helps You...

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