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# if in atwood machine m1=1.30kg and m2=2.80kg at t=o.m1 starts losing the mass at a rate of .200kg/s due to leakage.at what rate does the acc. of the container  changes at t=0sec and t=3 sec.when dos the acc. reach max. value

## 2 Answers

10 years ago

dm1/dt = -0.2kg/s           ( given)

dm1 = -0.2dt

integrating both sides

m1 = -0.2t + c

at t=0 , m1=1.3

therefore  ,  c=1.3

now , m1 = 1.3 - 0.2t ...............1                                      (mass of m1 at any time)

now accleration (a)= g(m2-m1)/(m2+m1)

a = g{m2 - (1.3-0.2t)}/{m2+(1.3-0.2t)}                        (putting m1 from eq 1)

a =g(1.5+0.2t)/(4.1-0.2t)           ...................2              (after putting value of m2)

da/dt = g(1.12)/{4.1-0.2t)2  ................3

at t=0 ,

da/dt = 0.652m/s2

at t=3

da/dt = 0.896m/s2

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