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An elevator and its load have a total mass of 800 kg. The elevator is originally moving downwards at 10 m $s^{- 1}$ it slows down to stop with constant acceleration in a distance of 25 m. Find the tension T in the supporting cable while the elevator is being brought to rest. (Take g = 10 m $s^{- 2}$ )

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8000 N

1600 N

9600 N

6400 N

answer is C.

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Detailed Solution

When the elevator is going down with a retardation T = m $(g + a)$

where a is here $\frac{u^{2}}{2 s} = \frac{100}{2 \times 25} = 2 m / s^{2} T = 800 (10 + 2) = 9600 N$

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