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(D) The apparent weight $N$ of a person in an elevator is given by $N = m(g + a)$,where $a$ is the upward acceleration.For case $A$ (Stands still): $a

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Accelerates upward with an acceleration equal to $4 \, m/s^2$

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(D) The apparent weight $N$ of a person in an elevator is given by $N = m(g + a)$,where $a$ is the upward acceleration.For case $A$ (Stands still): $a = 0$,so $N = mg = 40 	imes 9.8 = 392 \, N$.For case $B$ (Constant velocity): $a = 0$,so $N = mg = 392 \, N$.For case $C$ (Accelerates downward): $a = -4 \, m/s^2$,so $N = m(g - 4) = 40(9.8 - 4) = 40(5.8) = 232 \, N$.For case $D$ (Accelerates upward): $a = 4 \, m/s^2$,so $N = m(g + 4) = 40(9.8 + 4) = 40(13.8) = 552 \, N$.Comparing all cases,the force is greatest when the elevator accelerates upward.

$A$ boy having a mass equal to $40 \, kg$ is standing in an elevator. The force felt by the feet of the boy (apparent weight) will be greatest when the elevator $(g = 9.8 \, m/s^2)$:

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