91Ó°ÊÓ

In this case, only two forces will be acting on the woman; weight (W) and the normal force pushing up on her (N). Consider the upward direction to be positive and apply Newton’s second law to calculate the acceleration.

Given data:

The regular weight of the woman is $W$.

The weight of the woman when the elevator begins to move is $N=0.75W$.

The free body diagram of the woman is as follows:

The relation to calculate the vertical forces is given by:

$\begin{array}{c}\mathrm{Î\pounds }{F}_v=ma\\ N-W=ma\end{array}$

Here, m is the mass, a is the acceleration, and N is the normal force.

On plugging the values in the above relation, you get:

$\begin{array}{c}0.75W-W=ma\\ 0.75mg-mg=ma\\ a=-0.25\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ a=-2.45\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=-2.45\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the elevator. The negative sign indicates that the direction of acceleration is downward.