let the accletation of thread is a...
accleration of m1 , m2 & the point where thread touches the disk is same a...
accleration of center of mass of disc is zero....
now let tension bw m & m2 be T2  &  bw m & m1 be T1 then
at m2
 m2g - T2 = m2a                       ..........1
at m1
T1 - f = m1a                           ............2                             (f is friction force)
at pulley
        (T2 + m2g - T1)R = I(alfa)                               (net torque = I(alfa))
        T2 - T1 + m2g = ma/2              ..................3                      {I (disc) = mR2/2 , R is radius of disc}
adding 1 & 2 then substituting T2-T1 in eq 3 we get
a = 2(2m2g - f)/{2(m1+m2)+m}                                       {f is friction force  = km1g }
a = 2g(2m2 - km1)/{2(m1+m2)+m}
 
 
 
						

							
Let the tension in the string be T. The force of friction opposes motion of the mass m1. Therefore the work done by the frictional force will be negative. Let the acceleration of the system be 'a'. Now applying newton's second law of motion. 'f' is the force of friction acting between the mass m1 and the horizontal plane
m2g-T=m2a ....1
T-f=m1a ....2  where f=kN  (N is the normal reaction which is equal to th e weight of m1,  N=m1g)
solve the two equations and calculate the value of 'a' in terms of m1,m2,g and k
now we have the value of acc. a
the work done by the frictional force on the mass m1=Force of friction * displacement(S)
after time t, the displacement of the mass m1 is
S=1/2*a*t^2     (second equation of motion)
put the value of 'a' and calculate the value of S
now calculate the work done by the frictional force and put a negative sign.