Question

A man in lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3rd of the acceleration due to gravity.If the same man were in the same lift nwo moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity,then what is his apparent weight?

Answer 1

$$\begin{gathered} w=m\times \left(g+\frac{g}{3}\right) \\ w=\frac{4}{3}mg \\ m=\frac{3w}{4g} \\ now,whentheliftisgoingdownwithnaccelerationof\frac{g}{2} \\ then, \\ {W}_{app}=m\times g_{app} \\ =\frac{3w}{4g}\times \left(g-\frac{g}{2}\right) \\ =\frac{3w}{4g}\times \frac{g}{2} \\ {W}_{app}=\frac{3w}{8} \end{gathered}$$
$$\begin{gathered} w=m\times \left(g+\frac{g}{3}\right) \\ w=\frac{4}{3}mg \\ m=\frac{3w}{4g} \\ now,whentheliftisgoingdownwithnaccelerationof\frac{g}{2} \\ then, \\ {W}_{app}=m\times g_{app} \\ =\frac{3w}{4g}\times \left(g-\frac{g}{2}\right) \\ =\frac{3w}{4g}\times \frac{g}{2} \\ {W}_{app}=\frac{3w}{8} \end{gathered}$$