JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Work, Power & Energy

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.

Answer: 7

Solution

The lift starts at

u = 2

m s

^{- 1}

and then falls with constant acceleration

a = 2

m s

^{- 2}

through

s = 6

m. Its speed at the end follows from

v^{2} = u^{2} + 2 a s = 2^{2} + 2 (2) (6) = 4 + 24 = 28

^{2}

^{- 2}

. The kinetic energy at that instant is

K = \frac{1}{2} M v^{2} = \frac{1}{2} (500) (28) = 7000

= 7

kJ.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer

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