here tension in the string is counter balanced by weight of block of mass m1
so we can say
[tex]T = m_1g[/tex]
[tex]T = 0.254 \times 9.8 = 2.49 N[/tex]
now on the other side the block which is placed on the inclined plane
we can say that component of weight of the block and friction force is counter balanced by tension force
[tex]m_2g sin\theta + F_f = T [/tex]
now we can plug in all values to find the friction force
[tex]0.273 \times 9.8 sin38.382 + F_f = 2.49[/tex]
[tex]1.66 + F_f = 2.49[/tex]
[tex]F_f = 0.83 N[/tex]
so it will have 0.83 N force on it due to friction
now to find the friction coefficient
[tex]F_f = \mu \times F_n[/tex]
here we know that
[tex] F_n = m_2gcos\theta[/tex]
[tex]F_n = 0.273 \times 9.8 \times cos38.382 = 2.1 N[/tex]
now from above equation
[tex]0.83 = \mu \times 2.1[/tex]
[tex]\mu = 0.38[/tex]
so friction coefficient will be 0.38
