An elevator weighing 2000 kg falls with a downward acceleration of magnitude (1/2)g . What is the tension in the supporting cable?

I got the answer as 49000 N.

To find the tension in the supporting cable of the falling elevator, you can use Newton's second law of motion, which states that Force equals mass multiplied by acceleration (F = m * a).

In this case, the force in question is the tension in the cable, the mass is 2000 kg, and the acceleration is (1/2)g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Firstly, we need to find the acceleration of the elevator in terms of g:
Acceleration = (1/2)g = (1/2) * 9.8 m/s^2 = 4.9 m/s^2.

Next, calculate the tension in the cable:
Tension = mass * acceleration = 2000 kg * 4.9 m/s^2 = 9800 kg * m/s^2.

Finally, using the definition of a newton (N) as 1 kg * m/s^2, the tension in the cable is:
Tension = 9800 N.

Therefore, the tension in the supporting cable is 9800 N, not 49000 N.