Given data: Mass of Block A = 10Kg

Mass of Block B = 15Kg

Angle of inclination α = 300

Co-efficient of friction m = 0.2

Consider the motion of block B,

The acceleration of block B will be half the acceleration of the block A i.e. a/2,

M_{1}g – 2T = m_{1}(a/2)

15 X 9.81 – 2T = 15 (a/2)

147.15 – 2T = 7.5a ...(1)

Consider the motion of block B

T – W_{2}sin α – µW_{1}cos α = (W_{1}/g)a

T - m_{2}g sin α – 0.2m_{2} gcos α = m_{2}a

T – 10 X 9.81sin 300 – 0.2 X 10 X 9.81cos300 = 10a

T – 66.04 = 10a ...(2)

Adding equation (1) with 2 X equation (2)

147.15 – 2T + 2T – 132.08 = 7.5a + 20a a = 0.54 m/sec^{2}

Now velocity of the block after 10 sec,

Apply v = u + at

V = 0 + 0.54 X 10

V = 5.4m/sec