Laws of Motion - Result Question 28
28. The mass of a lift is $2000 kg$. When the tension in the supporting cable is $28000 N$, then its acceleration is:
[2009]
(a) $4 ms^{-2}$ upwards
(b) $4 ms^{-2}$ downwards
(c) $14 ms^{-2}$ upwards
(d) $30 ms^{-2}$ downwards
Show Answer
Answer:
Correct Answer: 28. (a)
Solution:
(a) Net force, $F=T-mg$
$2000 a=28000-20000=8000$
$a=\frac{8000}{2000}=4 ms^{-2} \uparrow$
