Chapter 4, Problem 65P

(a)

To determine

The reading of the scale when the elevator is ascending with the speed of $30\text{m}/\text{s}$ .

Answer to Problem 65P

The reading indicated by the scale is $19.6\text{N}$ .

Explanation of Solution

Given:

The mass of the block is $m=2.0\text{kg}$ .

Formula used:

The expression for the weight of the block is given by,

  $w=mg$

As the object is ascending at a constant speed, thus the reading of scale will be equal to the tension in the string and the expression for the tension of the string is given by,

  $T=w$

Calculation:

The weight of the block is calculated as,

  $\begin{array}{c}w=mg\\ =\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =19.6\text{N}\end{array}$

The tension of the string is calculated as,

  $\begin{array}{c}T=w\\ =19.6\text{N}\end{array}$

Conclusion:

Therefore, the reading indicated by the scale is $19.6\text{N}$ .

(b)

To determine

The reading of the scale when the elevator is descending with the speed of $30\text{m}/\text{s}$ .

Answer to Problem 65P

The reading indicated by the scale is $19.6\text{N}$ .

Explanation of Solution

Given:

The mass of the block is $m=2.0\text{kg}$ .

Formula used:

The expression for the weight of the block is given by,

  $w=mg$

As the object is descending at a constant speed, thus the reading of scale will be equal to the tension in the string and the expression for the tension of the string is given by,

  $T=w$

Calculation:

The weight of the block is calculated as,

  $\begin{array}{c}w=mg\\ =\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =19.6\text{N}\end{array}$

The tension of the string is calculated as,

  $\begin{array}{c}T=w\\ =19.6\text{N}\end{array}$

Conclusion:

Therefore, the reading indicated by the scale is $19.6\text{N}$ .

(c)

To determine

The reading of the scale when the elevator is ascending with the speed of $20\text{m}/\text{s}$ and is descending at the rate of $3.0\text{m}/{\text{s}}^2$ .

Answer to Problem 65P

The reading indicated by the scale is $25.6\text{N}$ .

Explanation of Solution

Given:

The mass of the block is $m=2.0\text{kg}$ .

Formula used:

Consider the upward acceleration of the block is $a=3\text{m}/{\text{s}}^2$ .

The expression for the reading of the scale is given by,

  $T=m\left(g+a\right)$

Calculation:

The value of the reading of the scale is calculated as,

  $\begin{array}{c}T=m\left(g+a\right)\\ =\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+3\text{m}/{\text{s}}^2\right)\\ =25.6\text{N}\end{array}$

Conclusion:

Therefore, the reading indicated by the scale is $25.6\text{N}$ .

(d)

To determine

The reading of the scale for the time interval $0<t<9.0\text{s}$ .

Answer to Problem 65P

The reading indicated by the scale for different time interval is $\begin{array}{l}0<t<5=19.6\text{N}\\ 5\le t<9=14.6\text{N}\end{array}$

Explanation of Solution

Given:

The mass of the block is $m=2.0\text{kg}$ .

Formula used:

The expression for the reading of the elevator during the first $5\text{s}$ is given by,

  $w=mg$

The block comes to rest in $4\text{s}$ from the speed of $10\text{m}/\text{s}$ , thus the acceleration is negative and is given by,

  $v=v_0-at$

The reading of the scale will be equal to the tension in the string and the expression for the tension in the string is given by,

  $T=m\left(g-a\right)$

Calculation:

The value of the reading of the scale during the first $5\text{s}$ is given by,

  $\begin{array}{c}w=mg\\ =\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =19.6\text{N}\end{array}$

The acceleration of the block after $4\text{s}$ and is given by,

  $\begin{array}{l}v=v_0-at\\ 0=10\text{m}/\text{s}-\left(4\text{s}\right)a\\ a=2.5\text{m}/{\text{s}}^2\end{array}$

The reading of the scale is calculated as,

  $\begin{array}{c}T=m\left(g-a\right)\\ =\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2-2.5\text{m}/{\text{s}}^2\right)\\ =14.6\text{N}\end{array}$

Thus, the reading of the scale for different time interval is given by,

  $\begin{array}{l}0<t<5=19.6\text{N}\\ 5\le t<9=14.6\text{N}\end{array}$

Conclusion:

Therefore, the reading indicated by the scale for different time interval is $\begin{array}{l}0<t<5=19.6\text{N}\\ 5\le t<9=14.6\text{N}\end{array}$