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Application of Newtons laws

Question

In the system of pulleys shown what should be the value of m1 (in gram) such that 100 g remains at rest.  (Take g=10m/s2 ) 

Moderate
Solution

Since 100 g is at rest,

lf m1 is going up at acceleration a, we use

2Tm1g=m1a   ……(1)

For 100 g to be at rest, acceleration of 200 g mass is 2a if m1 is going up at a

2T=0.2(2a)21=0.4aa=104=52m/s2

 From (1), we have 2(1)10m1=m152

2=10m1+5m124=25m1m1=425=0.16kg=160g



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