A 60 kg man is in a lift moving upward with acceleration 2.45 m/s². What is the apparent percentage change in his weight?

What the question is really asking

When a lift accelerates upward, the scale reading (normal reaction) is larger than the person’s true weight $mg$. We need the percentage increase in this scale reading compared to $mg$.

Relating apparent weight to acceleration

For a person of mass $m$ in a lift accelerating upward with acceleration $a$:

• Weight downward: $mg$
• Normal reaction upward (apparent weight): $N$

Net upward force: $$N - mg = ma$$ So, $$N = m(g+a)$$

Percentage change in weight

True weight is $W=mg$. The change is $\Delta W = N - W = m(g+a) - mg = ma$.

Percentage change: $$\%\text{ change} = \frac{\Delta W}{W}\times 100 = \frac{ma}{mg}\times 100 = \frac{a}{g}\times 100$$ Substitute $a=2.45\ \text{m/s}^2$ and $g=9.8\ \text{m/s}^2$: $$\%\text{ change} = \frac{2.45}{9.8}\times 100 = 25\%$$

Quick check

Because the lift accelerates upward, you should feel heavier, so the answer must be a positive percentage increase. 25% fits that expectation.