A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.

A mass `m_(1)` is connected by a weghtless cable of water, whose mass

1.	A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.
Answer» Correct Answer - `((m_(1)-m_(0)+kt)g+kv_(0))/((m_(1)+m_(0)-kt))` Writing equations: `m_(1)g-T=m_(1)a` ...(i) `T-mg=ma` ....(ii) Adding equations (i) and (ii) `g(m_(1)-m)=(m_(1)+m)a` ....(iii) As `-(dm)/(dt)=k` `-underset(m_(0))overset(m)intdm=underset(0)overset(t)intkdt` `m_(0)-m=kt+C` `m_(0)-kt=m` By putting value of m in equation (iii) we get acceleration

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A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.

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