Question

a person of mass m is standing in a lift find apparent weight
a) moving upward with uniform acceleration a

Answer 1

$$\begin{gathered} \mathrm{Apparent}\mathrm{weight}\mathrm{while}\mathrm{going}\mathrm{up}: \\ \mathrm{W}\text{'}=\mathrm{W}\left[1+\frac{\mathrm{a}}{\mathrm{g}}\right] \\ =\mathrm{mg}\left[1+\frac{\mathrm{a}}{\mathrm{g}}\right] \\ =\mathrm{mg}+\mathrm{ma} \\ \mathrm{where} \\ \mathrm{m}=\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{person} \\ \mathrm{g}=\mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity} \\ \mathrm{a}=\mathrm{Acceleration}\mathrm{of}\mathrm{the}\mathrm{lift} \end{gathered}$$