Q. 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 $ms^{- 1}$ is [Take $g = 10\ ms^{- 2}$]
A) 1 m
B) 2 m
C) $\frac{5}{32}\ $m
D) $\frac{5}{16}\ $m
Explanation
(d) $T = mg + ma$
2000g $= 1000g + 1000a$
or $a = g$
Direction is upward
Now, $0^{2} - {2.5}^{2} = - 2 \times 10 \times s$
or $s = \frac{2.5 \times 2.5}{20}$
$= \frac{625}{100 \times 20} = \frac{25}{80}m = \frac{5}{16}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.