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Two blocks mA=10 kg and mb=5 kg are connected with cord & pulley system as shown in fig

Determine the velocity of each block when system is started from rest & block B gets displacement by 2m. Take µk=0.2 between block A & horizontal surface.

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Consider length of string w.r.t. block A as XA &

Length of string w.r.t block B as XB

enter image description here

So, applying C.S.L.M. we get

2XA+XB=C

But as B moves down, XB increases & XA decreases

So, rewriting eqn

2XA+XB=C______(i) 

Differentiating w.r.t. t

2VA+VB=0............2VA=VB_____(ii) 

Differentiating once more w.r.t. t

2aA=aB____(iii) 

Applying NSL to block A

Fx=max  &  Fy=0T0.2(N)=10aA  N=10gT0.2(10g)=10aA_____(iv) 

Applying NSL to block A

Fy=may5gT=maB_____(v) 

Solving eqns (iii)(iv)&(v),we get

T=34.335NaA=1.4715m/s2aB=2.934m/s2

Now, after displacement of 2m, let velocity of B be VB

(VB)2=0+2aB(2)....VB=3.43m/s

Putting in eqn (ii), VA=1.715m/s

Velocity if A is 1.715m/s & Velocity of B is 3.43m/s

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