Question

A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 m s^–1 is - [Take g = 10 m s^–2]

• A. 1 m
• B. 2 m
• C. $\frac{5}{32}$m
• D. $\frac{5}{16}$m

Answer 1

Answer: (D)

$\frac{5}{16}$m

Answer 2

When lift is moving downward then  = m(g – a)

2000 g = 1000 (g – a), a = – g , Now v² = u² + 2as

0 = (2.5)² + 2(–g) × s or s = $\frac{(2.5)^{2}}{2 \times g}$