Question

A man of 60 kg weight is standing on a spring balance which is placed in a lift. If the lift ascends with a velocity of

20 m/sec, then observation of the balance will be :

Answer 1

answer :- the observation of the spring balance will be remain same e.g., 60kg .

explanation :- as you know spring balance observe normal reaction between contacting surface. it is effected only when lift accelerated or decelerated .
if lift is moving upward with acceleration a
then, observation of spring balance will be = m(g + a) , where m is mass of man

when lift is moving downward with acceleration a then, observation of spring balance will be = m(g - a) .

but when lift is moving upward or downard with constant velocity then, observation will be remain same.
hence, observation of man's weight is 60kg on spring balance.