Question

1. The coefficient of friction between the block of mass mı = 3.00 kg and the surface in the figure below is up = 0.455. The system starts from rest. What is the speed of the ball of mass m, = 5.00 kg when it has fallen a distance h = 1.80 m?

Answer 1


mg = 3kg the D.455 m = 5 kg va - Drawing free body diagram of me - AN= mig MxN = Fx M m > Tension (Frictional force) mig т Lapplying Newtons Law (II law) to both masses - for mass @ ie my which is accelerating with *. acceleration as considering direction of - mation as positive, - my a = T- Feo for mass my which is also moving with acceler- - -ation a - ma = mag- I Q

Rewriting both equation, m, a = To Fx ma = mag- I we know, F = Mr. 0 -0 k svormal force of coefficient of - Kinetic friction. mass mi f - Fx = Mc Mag. ego becomes - ma = TE Mk meg Soling for ST= ma + Mx mig substituting I in egno Mz a = mag-( m, at eg m. g) Mz a = m2 g -ma - Mr mig mat m, a = mag- Mix mig acm, + m2) = my g - My mig a = (mag- Mx m, g) - mm2)

a = - Substituting, Mag- Me mig themet m = m2 = Mr a g = 3 kg 5 kg 0.455 9,8 m/s2 3 a = 5 x 9.8 – 0.455 x 38.9.8 - 3+5) = 35.623 = 4.45 29 m/s2 8 . 1 Q = 4.4529 m/s2] Speed can be calculated as object started from Rest, ie initial speed = 0, le U=0 from eam of motion : v2= 4?t zas - - > v2= 295 - we have a = -4.4529 m/s2 __S=1.80m ş v2 = 2 x 4.4529 81180 _ V= 4.0038 m/s which is nearly equal » [V= 4 m/s I Am to, o U=0
so the speed of ball of mass 5 kg when it has fallen a distance of 1.80 m is 4.0038 m/s which is nearly equal to 4.0 m/s.