- Home
- Standard 11
- Physics
5.Work, Energy, Power and Collision
medium
A lift of mass $M =500\,kg$ is descending with speed of $2\,ms ^{-1}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2\,ms ^{-2}$. The kinetic energy of the lift at the end of fall through to a distance of $6 m$ will be $...........kJ.$
A
$7$
B
$5$
C
$4$
D
$3$
(JEE MAIN-2023)
Solution
$v ^2 = u ^2+2 as$
$=2^2+2(2)(6)$
$=4+24=28$
$KE =\frac{1}{2} mv ^2$
$=\frac{1}{2}(500) 28$
$=7000\,J$
$=7\,kJ$
Standard 11
Physics
