Chapter 5, Problem 19P

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

a. Before the elevator starts moving?

b. While the elevator is speeding up?

c. After the elevator reaches its cruising speed?

To determine

To find: The apparent weight of the passenger before the elevator starts moving.

Answer to Problem 19P

Solution: The apparent weight of the passenger before the elevator starts moving is $590\text{N}$.

Explanation of Solution

Given data:

Mass of the passenger $\left(m\right)$ is 60 kg.

Cruising speed of the elevator is $10\frac{\text{m}}{\text{s}}$.

Time taken by the elevator to reach its cruising speed is 4.0 s.

Formula used:

The expression for apparent weight of the passenger before the elevator starts moving is as follows:

$w_{\text{app}}=m\left(g+a_y\right)$ (1)

Here,

$m$ is the mass of the passenger.

$g$ is the acceleration due to gravity, which is $9.8\frac{\text{m}}{{\text{s}}^2}$, and

$a_y$ is the acceleration of the passenger.

Two forces act on the passenger along a single vertical line. They are the downward pull of the gravity and the upward push of the elevator.

The elevator is at rest and so is the passenger before the elevator starts moving up. Therefore, the acceleration of the passenger before the elevator starts moving is $0\frac{\text{m}}{{\text{s}}^2}$.

$a_y=0\frac{\text{m}}{{\text{s}}^2}$

Calculation of apparent weight of the passenger before the elevator starts moving:

Substitute 60 kg for $m$, $9.8\frac{\text{m}}{{\text{s}}^2}$ for $g$, and $0\frac{\text{m}}{{\text{s}}^2}$ for $a_y$ in Equation (1).

$\begin{array}{c}{w}_{\text{app}}=\left(60\text{kg}\right)\left(9.8\frac{\text{m}}{{\text{s}}^2}+0\frac{\text{m}}{{\text{s}}^2}\right)\\ =\left(60\text{kg}\right)\left(9.8\frac{\text{m}}{{\text{s}}^2}\right)\\ =588\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\\ \simeq 590\text{N}\left\{\because 1\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}=1\text{N}\right\}\end{array}$

Conclusion:

Thus, the apparent weight of the passenger before the elevator starts moving is $590\text{N}$.

To determine

To find: The apparent weight of the passenger while the elevator is speeding up.

Answer to Problem 19P

Solution: The apparent weight of the passenger while the elevator is speeding up is $740\text{N}$.

Explanation of Solution

The expression to find the acceleration is as follows:

$a_y=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}$ (2)

Here,

$v_{\text{i}}$ is the initial speed of the elevator.

$v_{\text{f}}$ is the final speed of the elevator (cruising speed), and

$\Delta t$ is the time taken by the elevator to reach its cruising speed.

The speed of the elevator initially is $0\frac{\text{m}}{\text{s}}$ and gradually reaches its cruising speed of $10\frac{\text{m}}{\text{s}}$ in 4 seconds.

Therefore, the initial speed of the elevator is $0\frac{\text{m}}{\text{s}}$.

$v_{\text{i}}=0\frac{\text{m}}{\text{s}}$

And, the final speed of the elevator is $10\frac{\text{m}}{\text{s}}$.

$v_{\text{f}}=10\frac{\text{m}}{\text{s}}$

Calculation of acceleration:

Substitute $0\frac{\text{m}}{\text{s}}$ for $v_{\text{i}}$, $10\frac{\text{m}}{\text{s}}$ for $v_{\text{f}}$, and 4.0 s for $\Delta t$ in Equation (2).

$\begin{array}{c}{a}_y=\frac{10\frac{\text{m}}{\text{s}}-0\frac{\text{m}}{\text{s}}}{4.0\text{s}}\\ =\frac{10\frac{\text{m}}{\text{s}}}{4.0\text{s}}\\ =2.5\frac{\text{m}}{{\text{s}}^2}\end{array}$

Calculation of apparent weight of the passenger, while the elevator is speeding up:

Substitute 60 kg for $m$, $9.8\frac{\text{m}}{{\text{s}}^2}$ for $g$, and $2.5\frac{\text{m}}{{\text{s}}^2}$ for $a_y$ in Equation (1).

$\begin{array}{c}{w}_{\text{app}}=\left(60\text{kg}\right)\left(9.8\frac{\text{m}}{{\text{s}}^2}+2.5\frac{\text{m}}{{\text{s}}^2}\right)\\ =\left(60\text{kg}\right)\left(12.3\frac{\text{m}}{{\text{s}}^2}\right)\\ =738\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\\ \simeq 740\text{N}\end{array}$

Conclusion:

Thus, the apparent weight of the passenger while the elevator is speeding up is $740\text{N}$.

To determine

To find: The apparent weight of the passenger after the elevator reaches its cruising speed.

Answer to Problem 19P

Solution:

The apparent weight of the passenger after the elevator reaches its cruising speed is $590\text{N}$.

Explanation of Solution

As the elevator does not accelerate anymore after it reaches its cruising speed, the acceleration of the passenger is $0\frac{\text{m}}{{\text{s}}^2}$.

$a_y=0\frac{\text{m}}{{\text{s}}^2}$

Calculation of apparent weight of the passenger after the elevator reaches its cruising speed:

Substitute 60 kg for $m$, $9.8\frac{\text{m}}{{\text{s}}^2}$ for $g$, and $0\frac{\text{m}}{{\text{s}}^2}$ for $a_y$ in Equation (1).

$\begin{array}{c}{w}_{\text{app}}=\left(60\text{kg}\right)\left(9.8\frac{\text{m}}{{\text{s}}^2}+0\frac{\text{m}}{{\text{s}}^2}\right)\\ =\left(60\text{kg}\right)\left(9.8\frac{\text{m}}{{\text{s}}^2}\right)\\ =588\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\\ \simeq 590\text{N}\end{array}$

Conclusion:

Thus, the apparent weight of the passenger after the elevator reaches its cruising speed is $590\text{N}$.