It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.

Question

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger’s apparent weight before the elevator starts moving?
2. What is the passenger’s apparent weight whilethe elevator is speeding up?
3. What is the passenger’s apparent weight afterthe elevator reaches its cruising speed?

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Philomena 4 years 2021-07-13T12:06:07+00:00 1 Answers 822 views 0

Answers ( )

  1. giabao
    2
    2021-07-13T12:07:50+00:00
    Reply

    Answer:

    1. 588 N

    2. 738 N

    3. 588 N

    Explanation:

    time, t = 4 s

    initial velocity, u = 0

    final velocity, v = 10 m/s

    mass, m= 60 kg

    1.

    Weight of passenger before starts

    W =m g = 60 x 9.8 = 588 N

    2.

    When the elevator is speeding up

    v = u + a t

    10 = 0 + a x 4

    a = 2.5 m/s2

    Now the weight is

    W’ = m (a + g) = 60 (9.8 + 2.5) = 738 N

    3.

    When he reaches the cruising speed, the weight is

    W = 588 N

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