Two blocks of mass m a = 5kg and m b =7kg are connected by a cable negligible ma

Question

Two blocks of mass ma = 5kg and mb=7kg are connected by a cable negligible mass on a pulley. The pulley rotates on a frictionless axle, the cable doesnt 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 of theta=30 degrees above the ground and there is a coeffcient of friction uk=0.12 between block A and the ramp. Block B sits a distance d=3cm 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.

Explanation / Answer

Let

m1 = 5 kg

m2 = 7 kg

mp = 10 kg

r = 4 cm = 0.04 m

moment of inertia of pulley, I = mp*r^2

= 10*0.04^2

= 0.016 kg.m^2

Let T1 and T2 are the tensions in the strings connected to m1 and m2.

let a is the acceleration of the two blocks.

Net force acting on m2,

Fnet2 = m2*g - T2

m2*a = m2*g - T2

T2 = m2*g - m2*a ---(1)

Net force acting on m1,

Fnet1 = T1 - m1*g*sin(30) - mue_k*m1*g*cos(30)

m1*a = T1 - m1*g*sin(30) - mue_k*m1*g*cos(30)

T1 = m1*a + m1*g*sin(30) + mue_k*m1*g*cos(30) ---(2)

Net torque acting on pulley, Tnet = T2*r - T1*r

I*alfa = T2*r - T1*r

mp*r^2*a/r = T2*r - T1*r

mp*a = T2 - T1

mp*a = m2*g - m2*a - m1*a - m1*g*sin(30) - mue_k*m1*g*cos(30)

a*(m1+m2+mp) = m2*g - m1*g*sin(30) - mue_k*m1*g*cos(30)

a = (m2*g - m1*g*sin(30) - mue_k*m1*g*cos(30))/(m1+m2+mp)

= (7*9.8 - 5*9.8*sin(30) - 0.12*5*9.8*cos(30))/(5+7+10)

= 1.77 m/s

Apply kinematic equation, v^2 - u^2 = 2*a*d

v^2 - 0^2 = 2*a*d

v = sqrt(2*a*d)

so, speed of block B when itbstrike the ground, v = sqrt(2*a*d)

= sqrt(2*1.77*3)

= 3.259 m/s <<<<<<<<-------Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.