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.

Question

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)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

v 2 = u 2+2 as =22+2(2)(6) =4+24=28 KE =(1/2) mv 2 =(1/2)(500) 28 =7000 J =7 kJ

Q. A lift of mass $M = 500 k g$ is descending with speed of $2 m s^{- 1}$ . Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 m s^{- 2}$ . The kinetic energy of the lift at the end of fall through to a distance of $6 m$ will be ___ $k J$ .

$v^{2} = u^{2} + 2 a s$
$= 2^{2} + 2 (2) (6)$
$= 4 + 24 = 28$
$K E = \frac{1}{2} m v^{2}$
$= \frac{1}{2} (500) 28$
$= 7000 J$
$= 7 k J$

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

v2=u2+2as =22+2(2)(6) =4+24=28 KE=21​mv2 =21​(500)28 =7000J =7kJ

Q. A lift of mass $M = 500 k g$ is descending with speed of $2 m s^{- 1}$ . Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 m s^{- 2}$ . The kinetic energy of the lift at the end of fall through to a distance of $6 m$ will be ___ $k J$ .

$v^{2} = u^{2} + 2 a s$
$= 2^{2} + 2 (2) (6)$
$= 4 + 24 = 28$
$K E = \frac{1}{2} m v^{2}$
$= \frac{1}{2} (500) 28$
$= 7000 J$
$= 7 k J$