Question

Two blocks of mass mA = 5kg and mB = 7kg are connected by a cable of negligible mass on a pulley. The

pulley rotates on a frictionless axle, the cable doesn’t slip on the pulley, it has a radius r = 4cm and a mass mp = 10g. Block A is on an inclined ramp at an angle θ = 30◦ above the ground and there is a coefficent of

friction μk = 0.12 between block A and the ramp. Block B sits a distance d = 3m from the ground. The system is released from rest and block B descends. Calculate the speed of block B at the instant just before it hits the ground

Answer 1

I = moment of inertia of pulley = (0.5) m_p r² = (0.5) (0.01) (0.04)² = 8 x 10^-6 kgm²

For Block A :

Perpendicular to incline force equation is given as

F_n = m_Ag Cos30

frictional force is given as

f_k = u_k F_n = u_k m_Ag Cos30

from the force diagram , force equation for block A is given as

T₁ - m_A g Sin30 - f_k = m_Aa

T₁ - 5 x 9.8 Sin30 - 0.12 x 5 x 9.8 Cos30 = 5 a

T₁ - 29.6 = 5 a         

T₁ = 29.6 + 5a                   Eq-1

For Block B :

force equation is given as

m_Bg - T₂ = m_B a

7 x 9.8 - T₂ = 7a

T₂ = 68.6 - 7a                             Eq-2

Torque equation for the pulley is given as

(T₂ - T₁) r = Ia/r

(68.6 - 7a - 29.6 - 5a ) = (8 x 10^-6) a/(0.04)²

a = 3.25 m/s²

Vi = initial speed of B

Vf = final speed

d = distance travelled = 3 m

using the equation

V_f² = V_i² + 2 a d

V_f² = 02 + 2 x 3.25 x 3

V_f = 4.42 m/s