Block A is of mass 30kg & block B is of mass 5kg. if Î¼ kinetic friction between the inclined & block A is 0.2, determine the speed of block A after it moves 1m down the plane, starting from rest.

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1 Answers
m=mass of A
M=mass of B
F=frictional force
first step: Find the constraint equation
2T.SaTSb=0
so 2AB=0
A=10 is given. Thus, B=20
Now the FBD of A block
mg sin 37  2T  F = ma_{A}
mg cos 37 = N = 30.g.4/5 = 24g
Tension is internal force.. So work done by tension is 0
Work by friction is F.s = Î¼N.10 = 2.24=48g
Potential Energy change will be mg.10+Mg20 = 200g
Increase of Kinetic Energy+increase of Potential Energy + energy Loss=0
KE change = 20048 = 152
1/2mV_{a}^{2}+1/2MV_{b}^{2} = 152
30. V^{2} + 5. 4 V^{2} = 304
V^{2} = 152/25 =
i may have done some mistake sindhu.. cos did this one in too much jaldi jaldi.. But pls check the method!