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What is the passenger’s weight while the elevator is speeding up?

What is the passenger’s weight while the elevator is speeding up?

  • What is the passenger’s weight while the elevator is speeding up?
  • What is the passenger’s weight while the elevator is at rest?
  • What is the passenger’s weight while the elevator reaches the cruising speed?

While 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. 

This question aims to find the weight of a passenger when the elevator is speeding up. The time, speed, and mass are given to calculate the elevator’s speed.

Moreover, this question is based on the concepts of physics. It mainly deals with the dynamics which concern the body’s motion under the action of different forces. Therefore, we are calculating the weight of a passenger when he is in the elevator.

Expert Answer

 
The weight of a passenger can be calculated as:

mass = $m = 60 kg$

time = $t = 4 s $

final velocity = $v_2 = 10 m/s$

acceleration of the elevator = $g = 9.81 m /s^2$

a) What is the passenger’s weight while the elevator speeds up?

Since we know that:

\[ v_2 = v_1 + at \]

When the elevator is at rest initial velocity is:

\[ v_1 = 0 \]

Therefore,

\[ v_2 = at \]

\[ a = \dfrac{v_2}{t} \]

\[ = \dfrac{10m/s^2}{4s} \]

\[ = 2.5 m/s^2 \]

Therefore, the weight of the passenger will be:

\[ W = m (a + g) \]

\[ = 60 kg . ( 2.5 m s^{-2} + 9.81 m s^{-2}) \]

\[ W = 738.6 N \]

b) What is the passenger’s weight while the elevator is at rest?

\[W = mg\]

\[ W = (60 kg) (9.8 ms^ {-2}) \]

\[ W = 588.6 N \]

c) What is the passenger’s weight while the elevator reaches the cruising speed?

With the maximum speed, the elevator acceleration become uniform. Therefore,

\[ a = 0 \]

\[ W = m(g + a) = mg \]

\[ W = (60 kg)(9.8 m s^{-2}) \]

\[ W = 588.6 N \]

Numerical Results

a) Passenger’s weight while the elevator is speeding up is:

\[W = 738.6 N\]

b) Passenger’s weight while the elevator is at rest:

\[W = 588.6 N\]

c) Passenger’s weight while the elevator reaches the cruising speed is:

\[W = 588.6 N\]
 

Example

A model airplane with a mass of 0.750 kg flies in a horizontal circle at the end of a 60.0 m control wire, with a speed of 35.0 m/s. Compute the tension in the wire if it makes a constant angle of 20.0° with the horizontal.

Solution

The tension in the wire can be calculated as:

\[F = T + mg \sin (\theta)\]

\[ ma = T + mg \sin ( \theta ); \text{ since F }= ma\]

\[\dfrac{mv^2}{d} = T + mg \sin (\theta); \text{ since a } = \dfrac{v^2}{d}\]

Therefore,

\[T = \dfrac{(0.75)(35)^2}{60} – (.75)(9.8)\sin(20)\]

\[T = 12.8 N\]

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