A 77-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.8

Question

A 77-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.87 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.8 s, and then comes to rest. What does the spring scale register in each of the following time intervals

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Thanh Thu 3 months 2021-08-19T04:43:19+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-08-19T04:44:40+00:00

    Answer:

    Explanation:

    From the question, we can see that the elevator is at a constant speed, invariably, acceleration is zero. And thus, there is no external force acting in the man, asides from force of gravitation acting on the man between those 5s. Now, we solve the question using newton’s second law of motion.

    The second law of motion states that, “the acceleration of an object is dependent upon two variables – the net force acting upon the object and the mass of the object.”

    Summation Fy = ma(y)

    F(n) – F(g) = 0

    F(n) – mg = 0

    F(n) = mg

    F(n) = 77*9.81

    F(n) = 755.37N

    0
    2021-08-19T04:44:51+00:00

    Answer:

    Explanation:

    Given:

    Mass, M = 77 kg

    U = 0 m/s

    g = 9.81 m/s^2

    A.

    Since the elevator is not moving, no change in velocity, a = 0 m/s^2

    Total force = N – Fg

    = (M × g) – (M × a)

    The only force acting is the normal force, N = M × g

    = 77 × 9.81

    = 755.37 N

    = 0.755 kN.

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