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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. determine the passenger's weight, before the elevator starts moving, while the elevator is speeding up, and after the elevator reaches its cruising speed

Respuesta :

Refer to the diagram shown below.

State A: Before the elevator starts moving.
Initially, the weight of the 60 kg passenger is
(60 kg)*(9.8 m/s²) = 588 N

State A to B: Elevator accelerates
The acceleration between states A and B is
a = (10 m/s)/(4 s) = 2.5 m/s²
The inertial force induced on the passenger is
(60 kg)*(2.5 m/s²) = 150 N
The effective weight of the passenger is
588 + 150 = 738 N

After state B: Dynamic equilibrium
After state B, the elevator moves at constant speed and the passenger is in dynamic equilibrium.
Therefore the passenger's weight is  588 N.

Answer:
Weight = 588 N, before the elevator starts;
             = 738 N, while the elevator is speeding up;
             =  588 N, when the elevator reaches cruising speed.
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johanrusli

The passenger's weight :

before the elevator starts moving → 588 N

while the elevator is speeding up → 738 N

after the elevator reaches its cruising speed → 588 N

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

elapsed time = t = 4.0 s

cruising speed = v = 10 m/s

initial speed = u = 0 m/s

mass of passenger = m = 60 kg

Asked:

weight of passenger = ?

Solution:

Firstly , we will calculate the acceleration of elevator to reach its cruising speed :

[tex]a = ( v - u ) \div t[/tex]

[tex]a = ( 10 - 0 ) \div 4.0[/tex]

[tex]a = 10 \div 4.0[/tex]

[tex]\boxed{a = 2.5 \texttt{ m/s}^2}[/tex]

[tex]\texttt{ }[/tex]

Next , we will use Newton's Law to solve this problem as follows:

→ Before the elevator starts moving

[tex]\Sigma F = ma[/tex]

[tex]N - w = m(0)[/tex]

[tex]N - w = 0[/tex]

[tex]N = w[/tex]

[tex]N = mg [/tex]

[tex]N = 60 \times 9.8[/tex]

[tex]\boxed{N = 588 \texttt{ N}}[/tex]

[tex]\texttt{ }[/tex]

→ While the elevator is speeding up

[tex]\Sigma F = ma[/tex]

[tex]N - mg = ma[/tex]

[tex]N = m(g + a)[/tex]

[tex]N = 60(9.8 + 2.5)[/tex]

[tex]\boxed{N = 738 \texttt{ N}}[/tex]

[tex]\texttt{ }[/tex]

→ After the elevator reaches its cruising speed

[tex]\Sigma F = ma[/tex]

[tex]N - w = m(0)[/tex]

[tex]N - w = 0[/tex]

[tex]N = w[/tex]

[tex]N = mg [/tex]

[tex]N = 60 \times 9.8[/tex]

[tex]\boxed{N = 588 \texttt{ N}}[/tex]

[tex]\texttt{ }[/tex]

Learn more

  • Impacts of Gravity : https://brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
  • The Acceleration Due To Gravity : https://brainly.com/question/4189441
  • Newton's Law of Motion: https://brainly.com/question/10431582
  • Example of Newton's Law: https://brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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