Using the kinematic equation
s = vo*t + 1/2*g*t^2
you can see that the acceleration of the lift is 1/3*g.
The apparent acceleration of the mass inside the falling lift is
g' = (g -1/3*g) = 2g/3.
So the reaction of the floor on the mass is 2mg/3
a lift starting from rest moves downward with constant acceleration. It covers a distance s in time t, where s = (1/6)(g)(t^2). A box of mass m is on the floor of the lift. find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.
i got 5/6(mg) and 11/12(mg) using a= d/t^2 and v^2 = u^2 + 2as but the answer is supposed to be 2/3(mg)
what is wrong with these formulae? i dont think i made any calculation mistakes