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?

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.

dnji

step-by-step explanation:

dskfbipo

24

step-by-step explanation: