An elevator weighing 280 kg is lowered in the shaft at a uniform acceleration rate, in the first 10

An elevator weighing 280 kg is lowered in the shaft at a uniform acceleration rate, in the first 10 seconds it travels 35 meters. find the tension of the rope on which the elevator weighs.

The lift is affected by the force of gravity and the pulling force of the rope, these forces are directly opposite. According to Newton’s second law:
F = Ft – T; ma = mg – T; T = mg – ma = m (g – a), where m is the mass of the elevator (m = 280 kg), g is the acceleration of gravity (we take g = 9.8 m / s ^ 2), and is the acceleration of the elevator.
Since V0 = 0, and the motion is uniformly accelerated, then
h = (a * t ^ 2) / 2; a = 2h / t ^ 2.
Rope tension:
T = m (g – (2h / t ^ 2)) = 280 * (9.8- (2 * 35/10 ^ 2)) = 280 * (9.8-70 / 100) = 280 * 9.1 = 2548 N.
Answer. The rope tension is 2548 N.