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20 April, 18:18

A miners' cage of mass 420 kg contains 3 miners of total mass 280 kg. The cage

is lowered from rest by a cable. For the first 10 seconds the cage accelerates

uniformly and descends a distance of 75 m. What is the force in the cable during

the first 10 seconds?

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Answers (1)
  1. 20 April, 20:56
    0
    5817 Newtons.

    Step-by-step explanation:

    Total mass of the cage + the miners = 700 Kg which is a downward force of 700g N.

    The net downward force = 700g - T where T is the tension (force) in the cable. The g = acceleration due to gravity = 9.81 m s-2.

    We calculate the acceleration of the cage by using an equation of motion:

    Distance = ut + 1/2 a t^2 where u = initial velocity, t = time and a = acceleration:

    75 = 0 (t) + 1/2 a (10^2)

    50a = 75

    a = 1.5 m s-2.

    So using Newtons second law of motion

    Force = mass * acceleration:

    700*9.81 - T = 700 * 1.5

    T = 700 * 9.81 - 700*1.5

    = 5817 N.
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