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 \(...........kJ.\)
- A \(7\)
- B \(5\)
- C \(4\)
- D \(3\)
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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