The free body diagram of the given system is shown above. Let us consider that the 15kg block is moving downward with the acceleration a.

From figure (a)

T + ma − mg = 0

T + 15a − 15g = 0

⇒ T = 15g − 15a……. (i)

From the figure (c)

T₁ − mg − ma = 0

T₁ − 5g − 5a = 0

⇒ T₁ = 5g + 5a ………. (ii)

From figure (b)

T = (T₁ + 5a + μR) = 0

⇒ T − (5g + 5a + 5a +μR) = 0 ……. (iii)

(where R = μg)

From Equations (i) and (ii),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90

⇒ a = 3.6 m/s²

From Equation (iii),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10

T= 96 N in the left string.

From Equation (ii),
T₁ = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
T₁= 68 N in the right string.