1. Problem a) states: A lift and person with total mass 1200 kg descend with acceleration $\frac{g}{10} = 1$ m/s$^2$ downward. Find the tension in the cable. 2. The forces on the lift-person system are tension $T$ upward and weight $W = mg$ downward. 3. Using Newton's second law, $\sum F = ma$, taking downward as positive: $$ mg - T = ma $$ where $m=1200$ kg, $g=10$ m/s$^2$, $a=1$ m/s$^2$. 4. Substitute values: $$ 1200 \times 10 - T = 1200 \times 1 $$ $$ 12000 - T = 1200 $$ 5. Solve for $T$: $$ T = 12000 - 1200 = 10800 \text{ N} $$ 6. Problem b) states: The person has mass $M$ kg. The reaction force from the floor on the person is 638 N. Find $M$. 7. Forces on the person: weight $Mg$ downward, floor reaction $R=638$ N upward. 8. The person accelerates downward at $a=1$ m/s$^2$, so: $$ Mg - R = Ma $$ 9. Rearranged: $$ Mg - Ma = R $$ $$ M(g - a) = R $$ 10. Substitute $g=10$, $a=1$, $R=638$: $$ M(10 - 1) = 638 $$ $$ 9M = 638 $$ 11. Solve for $M$: $$ M = \frac{638}{9} \approx 70.89 \text{ kg} $$ Final answers: - a) Tension $T = 10800$ N - b) Mass $M \approx 70.89$ kg