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