Chapter 4, Problem 52A

(a)

To determine

The force exerted by a scale in an elevator on Earth on a $53\text{kg}$ person provided the elevator moves up with constant speed.

Answer to Problem 52A

The force exerted by the scale in an elevator on the person is $\text{5}\text{.2}\times {10}^2\text{N}$ .

Explanation of Solution

Given:

The elevator moves at a constant speed.

The mass $\left(m\right)$ of the person is $53\text{kg}$ .

Consider the value of acceleration due to gravity $\left(g\right)$ is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a person stands on a scale in an elevator moving upward.

The force $\left(F_{\text{scale}}\right)$ exerted by the scale on the person is given by the relation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=ma+mg\\ F_{\text{scale}}=m\left(a+g\right)\end{array}$

Here, $F_{\text{net}}$ is the net force acting on the person, $m$ is the mass of the person, $a$ is the acceleration of the person, and $g$ is the acceleration due to gravity.

Calculation:

Consider the elevator is moving up with constant speed.

The acceleration $\left(a\right)$ of the person inside the elevator is zero as the elevator is moving up with constant speed.

Find the value of the force exerted by the scale in an elevator on the person as follows:

  $\begin{array}{c}{F}_{\text{scale}}=m\left(a+g\right)\\ =53\times \left(0+9.80\right)\\ =53\times 9.80\\ =519.93\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\end{array}$

  $\begin{array}{c}{F}_{\text{scale}}=519.4\text{N}\\ =\text{5}\text{.2}\times {10}^2\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the scale in an elevator on the person is $\text{5}\text{.2}\times {10}^2\text{N}$ .

(b)

To determine

The force exerted by a scale in an elevator on Earth on a $53\text{kg}$ person, provided the elevator slows at $2{\text{m/s}}^2$ while moving upward.

Answer to Problem 52A

The force exerted by the scale in an elevator on the person is $4.1\times {10}^2\text{N}$ .

Explanation of Solution

Given info:

The acceleration of the elevator is $-2{\text{m/s}}^2$ (the negative sign represents the slowing down of the elevator).

The mass $\left(m\right)$ of the person is $53\text{kg}$ .

Consider the value of acceleration due to gravity $\left(g\right)$ is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a person stands on a scale in an elevator moving upward.

The force $\left(F_{\text{scale}}\right)$ exerted by the scale on the person is given by the relation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=ma+mg\\ F_{\text{scale}}=m\left(a+g\right)\end{array}$

Here, $F_{\text{net}}$ is the net force acting on the person, $m$ is the mass of the person, $a$ is the acceleration of the person, and $g$ is the acceleration due to gravity.

Calculation:

Consider the elevator is moving upward and it slows at $2{\text{m/s}}^2$ .

The acceleration $\left(a\right)$ of the person inside the elevator is $-2{\text{m/s}}^2$ .

Find the value of the force exerted by the scale in an elevator on Earth on the person as follows:

  $\begin{array}{c}{F}_{\text{scale}}=m\left(a+g\right)\\ =53\left(-2+9.80\right)\\ =53\times 7.80\\ =413.4\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =4.1\times {10}^2\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the scale in an elevator on the person is $4.1\times {10}^2\text{N}$ .

(c)

To determine

The force exerted by a scale in an elevator on Earth on a $53\text{kg}$ person, provided the elevator speeds up at $2{\text{m/s}}^2$ while moving downward.

Answer to Problem 52A

The force exerted by the scale in an elevator on the person is $4.1\times {10}^2\text{N}$ .

Explanation of Solution

Given info:

The acceleration of the elevator is

  $+2{\text{m/s}}^2\left(\text{the}\text{positive}\text{sign}\text{shows}\text{elevator}\text{speeds up}\right)$ .

The mass $\left(m\right)$ of the person is $53\text{kg}$ .

Consider the value of acceleration due to gravity $\left(g\right)$ is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a person stands on a scale in an elevator moving upward.

The force $\left(F_{\text{scale}}\right)$ exerted by the scale on the person is given by the relation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=ma+mg\\ F_{\text{scale}}=m\left(a+g\right)\end{array}$

Here, $F_{\text{net}}$ is the net force acting on the person, $m$ is the mass of the person, $a$ is the acceleration of the person, and $g$ is the acceleration due to gravity.

Calculation:

Consider the elevator is moving downward and it speeds up at w $2{\text{m/s}}^2$ .

The acceleration $\left(a\right)$ of the person inside the elevator is $-2{\text{m/s}}^2$ .

Find the value of the force exerted by the scale in an elevator on Earth on the person as follows:

  $\begin{array}{c}{F}_{\text{scale}}=m\left(a+g\right)\\ =53\left(-2+9.80\right)\\ =53\times 7.80\\ =413.4\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =4.1\times {10}^2\text{N}\end{array}$

  $4.1\times {10}^2\text{N}$ .

Conclusion:

Thus, the force exerted by the scale in an elevator on the person is $4.1\times {10}^2\text{N}$ .

(d)

To determine

The force exerted by a scale in an elevator on Earth on a $53\text{kg}$ person provided the elevator moves down with constant speed.

Answer to Problem 52A

The force exerted by the scale in an elevator on the person is $\text{5}\text{.2}\times {10}^2\text{N}$ .

Explanation of Solution

Given info:

The elevator moves down at a constant speed.

The mass $\left(m\right)$ of the person is $53\text{kg}$ .

Consider the value of acceleration due to gravity $\left(g\right)$ is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a person stands on a scale in an elevator moving downward.

The force $\left(F_{\text{scale}}\right)$ exerted by the scale on the person is given by the relation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=ma+mg\\ F_{\text{scale}}=m\left(a+g\right)\end{array}$

Here, $F_{\text{net}}$ is the net force acting on the person, $m$ is the mass of the person, $a$ is the acceleration of the person, and $g$ is the acceleration due to gravity.

Calculation:

Consider the elevator is moving up with constant speed.

The acceleration $\left(a\right)$ of the person inside the elevator is zero as the elevator is moving down with constant speed.

Find the value of the force exerted by the scale in an elevator on the person as follows:

  $\begin{array}{c}{F}_{\text{scale}}=m\left(a+g\right)\\ =53\times \left(0+9.80\right)\\ =53\times 9.80\\ =519.93\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\end{array}$

  $\begin{array}{c}{F}_{\text{scale}}=519.4\text{N}\\ =\text{5}\text{.2}\times {10}^2\text{N}\end{array}$

Hence, the force exerted by the scale in an elevator on the person is $\text{5}\text{.2}\times {10}^2\text{N}$ .

Conclusion:

Thus, the force exerted by the scale in an elevator on the person is $\text{5}\text{.2}\times {10}^2\text{N}$ .

(e)

To determine

The force exerted by a scale in an elevator on Earth on a $53\text{kg}$ person provided the elevator moves down with a constant acceleration of $2.5{\text{m/s}}^2$ .

Answer to Problem 52A

The force exerted by the scale in an elevator on the person is $6.5\times {10}^2\text{N}$ .

Explanation of Solution

Given info:

The elevator moves down at a constant speed.

The mass $\left(m\right)$ of the person is $53\text{kg}$ .

Consider the value of acceleration due to gravity $\left(g\right)$ is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a person stands on a scale in an elevator moving downward.

The force $\left(F_{\text{scale}}\right)$ exerted by the scale on the person is given by the relation:

  $\begin{array}{l}{F}_{\text{scale}}=F_{\text{net}}+F_{\text{g}}\\ F_{\text{scale}}=ma+mg\\ F_{\text{scale}}=m\left(a+g\right)\end{array}$

Here, $F_{\text{net}}$ is the net force acting on the person, $m$ is the mass of the person, $a$ is the acceleration of the person, and $g$ is the acceleration due to gravity.

Calculation:

Consider the elevator is moving up with constant speed.

The constant acceleration $\left(a\right)$ of the person inside the elevator is $2.5{\text{m/s}}^2$ as the elevator is moving down.

Find the value of the force exerted by the scale in an elevator on the person as follows:

  $\begin{array}{c}{F}_{\text{scale}}=m\left(a+g\right)\\ =53\times \left(2.5+9.80\right)\\ =53\times 12.3\\ =651.9\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\end{array}$

  $\begin{array}{c}{F}_{\text{scale}}=651.9\text{N}\\ =6.5\times {10}^2\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the scale in an elevator on the person is $6.5\times {10}^2\text{N}$ .