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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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 \(...........kJ.\)

  1. A \(7\)
  2. B \(5\)
  3. C \(4\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(7\)

Step-by-step Solution

Detailed explanation

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