Question

A body of mass 50kg is hung by a spring balance in a lift. Calculate the reading of the balance when: (i) The lift is ascending with an acceleration of 2m/s2
. (ii) The lift is descending with a constant velocity of 2m/s. (iii) The lift is descending with an acceleration of 2m/s2
.

Answer 1

Answer:

Explanation: when the lift ascends 1)tension in the spring=

T-mg=ma 〰T=m(g+a)

T=50(10+2)=600N

Reading= 60 kg is 600 /10 m/s^-2

2) when the velocity is constant there is no acceleration so , the mass would be the same i.e. 50 kg.

3)Similarly, when the lift descends,

Mg -T =ma

T=m(g-a)

=50(10-2)

=400N

Reading=40 kg