Apparent Weight In An Accelerating Lift

The scale reading equals the normal force it exerts on the person, which by Newton's Second Law need not equal the true weight when the lift accelerates. Taking upward as positive, the forces on the person are the normal force $N$ up and gravity $mg$ down, giving $N-mg=ma$. Solving, $N=m(g+a)=60(10+2)=60\times 12=720\text{ N}$. The 600 N value is the true weight, which would be the reading only if the lift moved at constant velocity or stood still. The 480 N value, equal to $m(g-a)$, would apply for downward acceleration, not upward. The 120 N value mistakenly uses $ma$ alone, omitting the gravitational term. It is worth stressing that gravity itself has not changed; only the support force has, because the floor must now both hold the person up and accelerate them upward. This follows the NCERT treatment of apparent weight in non-uniformly moving frames, and the same equation predicts weightlessness, $N=0$, in free fall when $a=-g$. As a plausibility check, upward acceleration should make a person feel heavier, so a reading greater than the 600 N true weight is exactly what we expect, and the 20 percent excess matches the ratio $a/g=0.2$.