A lift of mass M =500,kg is descending with s | vedclass

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Question

$v ^2 = u ^2+2 as$

$=2^2+2(2)(6)$

$=4+24=28$

$KE =\frac{1}{2} mv ^2$

$=\frac{1}{2}(500) 28$

$=7000\,J$

$=7\,kJ$

Vedclass · Accepted Answer

$7$

Vedclass · Answer

$v ^2 = u ^2+2 as$

$=2^2+2(2)(6)$

$=4+24=28$

$KE =\frac{1}{2} mv ^2$

$=\frac{1}{2}(500) 28$

$=7000\,J$

$=7\,kJ$

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 $...........kJ.$