Question

Find the apparent weight of a man weight 49 Kg on earth where he is standing in a life which is irising with an acceleration of $1.2m/s^{2}ii$) going with the same acceleration iii)
falling freely the action gravity iv) going up down with uniform velocity . Given $g=9.8m/s^2$

Answer 1

Given that,

Acceleration $a=1.2m/s^2$

Mass $M=49kg$

Now, in the first case when the man is moving up wards in the lift, we will have apparent weight

$F_N=mg+ma$

$F_N=m(g+a)$

$F_N=49(9.8+1.2)$

$F_N=49\times 11$

$F_N=539N$

Now, in the second case when the man is moving downwards in the lift, we will have apparent weight

$F_N=mg-ma$

$F_N=m(g-a)$

$F_N=m(9.8-1.2)$

$F_N=m\left(8.6\right)$

$F_N=49\times 8.6$

$F_N=421.4N$

Now, when falling freely

The apparent weight

$F_N=mg-mg$

$F_N=0$

Now, when moving up and down with constant velocity

$a=0$

Now, apparent weight

$F_N=mg-ma$

$F_N=49\times 9.8$

$F_N=480.2N$

This is answers.