In: Physics
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
so, speed of block B when itbstrike the ground, v =
sqrt(2*a*d)
= sqrt(2*1.77*3)
= 3.26 m/s <<<<<<<<-------Answer