An elevator and its load have a total mass of $800 kg$ . If the elevator, originally moving downward at $10 m/s$ , is brought to rest with constant deceleration in a distance of $25 m$ , the tension in the supporting cable will be ( $g = 10 {m/s}^{2}$ )

8000 N

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6500 N

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11200 N

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9600 N

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Solution

The correct option is D $9600 N$
Given
$u = 10 m/s$
$s = 25 m$
Let deceleration downwards be $^{'} a^{'}$
$\Rightarrow v^{2} - u^{2} = - 2 a s$
$\Rightarrow 0 - 100 = - 2 a \times (25)$
$\Rightarrow a = 2 {m/s}^{2}$
A elevator is moving downward with $a = - 2 {m/s}^{2}$
So $T = m (g - a)$
$= 800 (10 - (- 2))$
$T = 9600 N$

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