Question

Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass m₁=6.20 kg is released from rest at a height h=4.20 m above the table.

 (a) Using the isolated system model, determine the speed of the object of mass m2 = 3.00 kg just as the 6.20-kg object hits the table. 

 (b) Find the maximum height above the table to which the 3.00-kg object rises.

Answer 1

Then ,

v^2 = 2*g*s

Then ,

m2= h+s

Solution then forle lavation is mig-T= mia Here T- Tension in the string T-mag=maa en acceleration will be same because both mass affach with same steeing, then both cavationis, (141-42) g = (mitma) a here mi=6.20kg , m2=3rooky ,g=9. C 6.20 – 3.0) 9.8 = (6020+ 3.00) a = 9.8 m/s finda, a = ( 6.20-3:00) 9.8 (6.20+3700) 3.41s so) la = 3.41 m/s² fhen determine speed of obsech means determine is va Ve=aan here h=4.20m, V?= 2*3.418 4.20 = 28.64 U-√28.64 = 5.35mils |u=5.35m/s) And

Portb) then my hits the table tension will become o. mimoves freely travelling 21,20m distance so mais, find ma, so, v²=2xgxs S=? yos (s. 35)= 2x9.8*5 2x9.8 x S so (S35)? - 2xg. 8 = lonom 1s=1046m) find manimum height of me, so, total height from table=hts so, ma=hts ma is, ma=4.20ht 1.46m ma=5.66m A I hope help you!!