Numerical 1: Apparent Weight in Lift

Given: Mass of person m = 60 kg, lift acceleration a = 2 m/$s^{2}$ (upward), g = 10 m/$s^{2}$

Step 1: Identify direction of acceleration relative to g. Lift accelerates upward → use W' = m(g + a)

Step 2: Substitute values with units. $W' = m(g + a) = 60\,\text{kg} \times (10\,\text{m/s}^2 + 2\,\text{m/s}^2)$

Step 3: Compute. $W' = 60\,\text{kg} \times 12\,\text{m/s}^2 = 720\,\text{kg·m/s}^2 = \boxed{720\,\text{N}}$

Check: W' > mg (600 N) ✓ — person feels heavier when lift goes up.

Free fall check (a = g = 10 m/$s^{2}$): $W' = m(g - g) = 60 \times 0 = \boxed{0\,\text{N}}$