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A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increases to 1050 n, then returns to normal. when the elevator slows to a stop at the 10th floor, his weight drops to 588 n, and then returns to normal. determine the acceleration at the beginning and end of the trip.

Respuesta :

Edufirst
1) at rest his weight is 840 N

=> 840N = mass * g => mass = 840 N / g = 840 N / 9.8 m/s^2 = 85.7 kg

2) as the elevator rises, his weight increases to 1050 N,

The reading of the scale is the norma force of it over the body of the person.

And the equation for the force is: Net force = mass * acceleration = normal force - weight at rest

=> mass * acceleration = 1050 N - 840 N = 210 N

acceleration = 210 N / mass = 210 N / 85.7 kg = 2.45 m/s^2 (upward)

3)  when the elevator slows to a stop at the 10th floor, his weight drops to 588 N

=>  mass * acceleration = 588 N - 840 N = - 252 N

=> acceleration = - 252 N / 85.71 kg = - 2.94 m / s^2 (downward)

Answer:

Acceleration at the beginning of the trip 2.45 m/s^2 upward
Acceleration at the end of the trip 2.94 m/s^2 downward

Acceleration at the beginning of the trip is [tex]2.45\; m/s^2[/tex]upward.

Acceleration at the end of the trip is [tex]2.94\; m/s^2[/tex] downward.

Explanation:

Given information:

The weight of student at rest[tex]=840\;N[/tex]

So, the mass [tex]m=840/9.8[/tex]

[tex]m=85.7\;\text{kg}[/tex]

As the elevator rises:

His weight increases to [tex]1050\; N[/tex]

Now, the equation of the force is:

[tex]\text{Net force} = \text {mass}\;\times\; \text{acceleration}=\text{normal\;force}-\text{weight at rest}[/tex]

[tex]m\times a =1050-840\\m\times a =210\; N\\a=210/m\\a=210/85.7\\a=2.45\;\text{m}/s^2}[/tex](upward)

Now , when the elevator stops at 10th floor

weight [tex]=588\;N[/tex]

[tex]m\times a =588-840\\a=-252/85.71\\a=-2.94\;\text{m}/s^2[/tex](downward)

Hence,

Acceleration at the beginning of the trip is [tex]2.45\; m/s^2[/tex]upward.

Acceleration is at the end of the trip  [tex]2.94\; m/s^2[/tex] downward.

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