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A 1000Kg lift is supported by a cable that can support by 2000Kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 m/s is (g = 10 m/s²)

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$S = 5 / 16 m$

$S = 3 / 16 m$

$S = 7 / 16 m$

None

answer is A.

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Detailed Solution

$T = m [g + a] \Rightarrow 2000 \times 10 = 1000 [10 + a] \Rightarrow 20 = 10 + a \Rightarrow a = 10 m / s^{2} v^{2} - u^{2} = 2 a s \Rightarrow s = \frac{u^{2}}{2 a} = \frac{2.5 \times 2.5}{2 \times 10} \Rightarrow s = \frac{5}{16} m$

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