please solve it brainliest really
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let the unknown acceleration be a
therefore net forces acting on the block of mass 1 kg is
T-m(g)=m(a)-------------(because the 1 kg block will move up)
T-1(10)=1(a)
T=10+a---------(1)
and the netforces acting on the block of mass 4kg
4g-2T=4(a)-----------(because it will move down)
putting value of T from eqn(1):
40-2(a+10)=4a
solving:
a=20/6=10/3------(2)
i took value of g as 10 .
therefore net forces acting on the block of mass 1 kg is
T-m(g)=m(a)-------------(because the 1 kg block will move up)
T-1(10)=1(a)
T=10+a---------(1)
and the netforces acting on the block of mass 4kg
4g-2T=4(a)-----------(because it will move down)
putting value of T from eqn(1):
40-2(a+10)=4a
solving:
a=20/6=10/3------(2)
i took value of g as 10 .
Aaryansh17:
actually there is a small mistake where i wrote net forces for the second body, instead of 4a , it should be 8a in the R.H.S as the force is 2Tupwards and hence the answer will be 5/2 or 2.5(approx)
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