Chapter 5, Problem 50P

What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves (a) upward with constant speed 5.0 m/s, (b) downward with constant speed 5.0 m/s, (c) with an upward acceleration 0.23 g, (d) with a downward acceleration 0.23 g_,and (e) in free fall?

To determine

The spring scale reading for the weight of awoman in an elevator when it moves upward with a constant speed of $5\text{m/s}$ .

Answer to Problem 50P

Solution:

The spring scale will read a value of $569\text{N}$ or $5\text{8}\text{kg}$ when it moves upward with a constant speed of $5\text{m/s}$ .

Explanation of Solution

Given:

The speed of the elevator, $v=5.0\text{m/s}$

Mass of the woman in the elevator, $m=\text{58}\text{kg}$

Formula Used:

  $F_{net}=F_n-F_g$

Where $F_n$ is the normal force

  $F_g$ is the actual force

Calculation:

  $F_{net}=F_n-F_g$

Substitute all the values and solve:

  $m\left(0\right)=F_n-58\times 9.81$

  $F_n=569\text{N}$

Conclusion:

Therefore, the spring scale will read a value of $569\text{N}$ or $5\text{8}\text{kg}$ when it moves upward with a constant speed of $5.0\text{m/s}$ .

To determine

The spring scale reading for the weight of a woman in an elevator when it moves downward with a constant speed of $5.0\text{m/s}$ .

Answer to Problem 50P

Solution:

The spring scale will read a value of $569\text{N}$ or $5\text{8}\text{kg}$ when it moves downward with a constant speed of $5.0\text{m/s}$ .

Explanation of Solution

Given:

The speed of the elevator, $5.0\text{m/s}$

Mass of the woman in the elevator, $5\text{8}\text{kg}$

Formula Used:

  $F_{net}=F_n-F_g$

Where $F_n$ is the normal force

  $F_g$ is the actual force

Calculation:

  $F_{net}=F_n-F_g$

Substitute all the values and solve:

  $m\left(0\right)=F_n-58\times 9.81$

  $F_n=569\text{N}$

Conclusion:

Therefore, the spring scale will read a value of $569\text{N}$ or $5\text{8}\text{kg}$ when it moves downward with a constant speed of $5.0\text{m/s}$ .

To determine

The spring scale reading for the weight of awoman in an elevator when it moves with an upward acceleration of $0.23g$ .

Answer to Problem 50P

Solution:

The spring scale will read a value of $\text{699}\text{.84}\text{N}$ or $77.34\text{kg}$ when it moves with an upward acceleration of $0.23g$ .

Explanation of Solution

Given:

Upward acceleration of the elevator, $a=0.23g{\text{ m/s}}^{\text{2}}$

Mass of the woman in the elevator, $5\text{8}\text{kg}$

Formula Used:

  $F_{net}=F_n-F_g$

Where $F_n$ is the normal force

  $F_g$ is the actual force

Calculation:

  $F_{net}=F_n-F_g$

Substitute all the values and solve:

  $58\left(0.23g\right)=F_n-58\times 9.81$

  $58\left(0.23\times 9.81\right)=F_n-58\times 9.81$

  $F_n=699.84\text{N}$

Conclusion:

Therefore, The spring scale will read a value of $\text{699}\text{.84}\text{N}$ or $77.34\text{kg}$ when it moves with an upward acceleration of $0.23\text{g}$ .

To determine

The spring scale reading for the weight of a woman in an elevator when it moves with a downward acceleration of $0.23\text{g}$ .

Answer to Problem 50P

Solution:

The spring scale will read a value of $438.1\text{1}\text{N}$ or $44.66\text{kg}$ when it moves with a downward acceleration of $0.23g$ .

Explanation of Solution

Given:

Downward acceleration of the elevator, $a=0.23g{\text{ m/s}}^{\text{2}}$

Mass of the woman in the elevator, $5\text{8}\text{kg}$

Formula Used:

  $F_{net}=F_n-F_g$

Where $F_n$ is the normal force

  $F_g$ is the actual force

Calculation:

  $F_{net}=F_n-F_g$

Substitute all the values and solve:

  $58\left(0.23g\right)=F_n-\left(-58\times 9.81\right)$

  $58\left(0.23\times 9.81\right)=F_n+58\times 9.81$

  $F_n=-438.11\text{N}$

Conclusion:

Therefore, the spring scale will read a value of $438.11\text{N}$ or $44.66\text{kg}$ when it moves with a downward acceleration of $0.23g$ .

To determine

The spring scale reading for the weight of awoman in an elevator when it freely falls.

Answer to Problem 50P

Solution:

The spring scale will read a value of $0\text{N}$ or $0\text{kg}$ when the elevator falls freely.

Explanation of Solution

Given:

The speed of the elevator, $v=5.0\text{m/s}$

Mass of the woman in the elevator, $5\text{8}\text{kg}$

Formula Used:

  $F_{net}=F_n-F_g$

Where $F_n$ is the normal force

  $F_g$ is the actual force

Calculation:

  $F_{net}=F_n-F_g$

For a free fall, the acceleration is equal to the value of $g$ .

Substitute all the values and solve:

  $58\left(9.81\right)=F_n-\left(-58\times 9.81\right)$

  $F_n=0\text{N}$

Conclusion:

Therefore, the spring scale will read a value of $0\text{N}$ or $0\text{kg}$ when the elevator falls freely.