Question
Question: The coefficient of friction \(\mu \) is equal to. a
Where
m1=2kgm2=4kg
Substituting the values in it will give,
F=(4+2)a
F=6a
From this we can find the frictional force acting on2kg,
That is, we can write that,
Fr=F−m1a
It is already given that,
m1=2kgm2=4kgF=0.5t
And also consider the value of tat the time of slipping be
t=12.5
Substituting the values in the above obtained equations will give,
F=(m1+m2)a……….. (1)
Fr=F−m1a………….. (2)
Rearranging and comparing will leads to,
a=(2+4)0.5×12=1ms−1
From the equation (2), we can write that, after substituting the value of acceleration in it,
Fr=0.5×12−2×1=4
Now the condition for slipping begins, which will give as the relation,
fr≤frmax
For simplifying or to find the answer, we can take the equality relation between them.
fr=frmax
That means we can write that,
4=μm1g
Substituting the values in it,
4=μ×2×10
Therefore after rearranging this equation, we will get the coefficient of friction,
That is,
μ=0.2
Hence the coefficient of friction has been obtained.
So, the correct answer is “Option C”.
Note: Generally, the coefficient of kinetic friction is only dependent on the nature of the material of the surface. It is not dependent on any other factors. For example, the relative speed between the surfaces and the surface area in contact.