A box of weight $W$ rests on platform of a lift. When the platform is moving upwards with acceleration $a$, the normal contact force of the platform on the box has magnitute $kW$. When the platform is moving downwards with acceleration $2a$, the box remains in contact with it. Find the normal contact force in terms of $k$ and $W$, and deduce that $k < \frac{3}{2}$.
How do you deduce the k in this question?
Sum of the forces equals to mass times acceleration. In the first case, the normal force on the object is $N_1$ acting upwards. You get $$ma=N-W=kW-W=(k-1)W$$ You therefore get $$a=\frac{(k-1)W}{m}$$ When the platform is moving downward, the weight is greater than the normal force $$m\cdot 2a=W-N_2\\N_2=W-m\cdot 2a=W-m\frac{2(k-1)W}{m}=(3-2k)W$$ Note that $N_2$ has to be positive, so this will get you the condition for $k$