Block A is of mass 30kg & block B is of mass 5kg. if μ 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
62m=mass of A
M=mass of B
F=frictional force
first step: Find the constraint equation
2T.Sa-TSb=0
so 2A-B=0
A=10 is given. Thus, B=20
Now the FBD of A block
mg sin 37 - 2T - F = maA
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 = μN.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 = 200-48 = 152
1/2mVa2+1/2MVb2 = 152
30. V2 + 5. 4 V2 = 304
V2 = 152/25 =
i may have done some mistake sindhu.. cos did this one in too much jaldi jaldi.. But pls check the method!
