Question

Q2)) A block of mass m= 4 kg slides down a frictionless plane inclined 30° to the horizontal . The block is connected to other block ( moving vertically) of mass m2= 1kg by massless string through a rough pulley (with moment of ineria I= 12 mR", m= 2kg and R=0.5 m) as shown in the fig. a) Find the acceleration of the blocks. b) Find the acceleration of the pulley. c) Find the tension in the string . 30

Answer 1

folutions solution: oma m2 UD equation of motion for me muig. sino - T = mia. -O air acceleration along the incline. equation of motion for pully. R(T1 - T2 = 1.x = l. a 1=1 M.R? T - Tempoh equation of motion for me W w2.9 = M2.a. Ta = mc.cgra). - 1

from ① & W T.-T2 = I'm a = (mig.sino- nua) - malg+a) mr.gising-mag - (mut m2).a. (m & mutme). a = (my.xino- m2).g. putting values. (² + 4+1) a= (u.sino-1) *9.8 la = 1.635 m/s² & angular acceleration of pulley. 10+ - HTS = 3:27 rod/sel] T, = mig.sino-mi a = wilgisino-a) TIE 4 (9.81. 8in3o-1.635) = 13.08 N Tz = malgra) T2 = 1 1 9.81+ 1.635) = 11.445 N