# A 150kg load lies at the bottom of the descending elevator car and presses on the floor with a force of 1300N.

A 150kg load lies at the bottom of the descending elevator car and presses on the floor with a force of 1300N. Determine the acceleration of the elevator.

Initial data: m (cargo weight) = 150 kg; F (force with which the load presses on the floor) = P (load weight) = 1300 N.
Reference data: g (acceleration due to gravity) = 10 m / s ^ 2.
Since the lift moves down, the weight of the load can be expressed using the formula: P = m * (g – a), whence a = g – P / m.
Let’s perform the calculation: a = 10 – 1300/150 = 1.33 m / s ^ 2.
Answer: The acceleration of the elevator is 1.33 m / s ^ 2.

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