Question

A 1000 kg lift is supported by a cable that can
support 2000 kg. The shortest distance in which
the lift can be stopped when it is descending with
a speed of 2.5 ms-t is [Take g = 10 ms-2]​

Answer 1

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Answer 2

Given:

Mass of a lift = 1000kg

A cable support mass = 2000kg

Speed = 2.5 ms⁻¹

g = 10 ms⁻²

To find:

Shortest distance

Solution:

Now, maximum supported weight by a cable = 2000kg

        Maximum possible tension = 2000×g

        Tension currently = 1000kg

therefore,

   tension currently × (g-a) = max. possible tension

   1000 × (g - a) = 2000×g

   1000×g - 1000×a = 2000×g

     $a = \frac{1000 \times g - 2000 \times g}{1000}$

   a = -g

now, for shortest distance,

formula used,

v² - u² = 2as

here, v = final velocity = 0ms⁻¹

         u = initial velocity = 2.5 ms⁻¹

         a = acceleration = -g = -10 ms⁻²

          s = shortest distance

so, by substituting all the values in the equation, we get

     (0)² - (2.5)² = 2 × -10 × s

       - (2.5)² = - 20s

            $s = \frac{(2.5)^2}{20}$

            s = 0.3125m

Therefore, shortest distance will be 0.3125m.