The mass of the elevator car with passengers is 800 kg. Determine the module and the direction

The mass of the elevator car with passengers is 800 kg. Determine the module and the direction of acceleration of the elevator, if the weight of the elevator car with passengers is 7040N during movement.

Data: m (mass of the elevator car with passengers) = 800 kg; P (weight of the elevator car with passengers at accelerated movement) = 7040 N.

Constants: g = 10 m / s2.

At accelerated movement, the weight of the elevator car with passengers: P = m * g ± m * a; “-” – the elevator car moves down and “+” – the car moves up.

7040 = 800 * 10 ± m * a;

7040 = 8000 ± m * a (the elevator car moves down).

7040 = 8000 – m * a.

m * a = 8000 – 7040 = 960.

a = 960/800 = 1.2 m / s2.

Answer: The elevator car moves downward with an acceleration of 1.2 m / s2.