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A lift of mass \mathrm{M=500 kg} is descending with speed of \mathrm{ 2 ms^{-1}}. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of \mathrm{ 2 ms^{-2}}. The kinetic energy of the lift at the end of fall through to a distance of \mathrm{ 6 m} will be _______\mathrm{ kJ.}

Option: 1

7


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Acceleration is constant,

\mathrm{v^{2}=u^{2}+2as}

\mathrm{v^{2}=2^{2}+2(2)(6)}

\mathrm{v^{2}=28}

\mathrm{\frac{1}{2}Mv^{2}=\frac{1}{2}\times 500\times 28}

\mathrm{KE=7KJ}

Posted by

Anam Khan

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