Chapter 6, Problem 4RQ

If both sacks in the preceding question are lifted their respective distances in the same time, how does the power required for each compare? How about for the case where the lighter sack is lifted its distance in half the time?

To determine

The power required to lift both the sacks and the power required to lift the lighter sack in half the time.

Answer to Problem 4RQ

The power required to lift both the sacks is same and the power required to lift the lighter sack in half the time is twice to the time required to lift the sack in time $t$.

Explanation of Solution

Given Info: Mass of sack in case (a) is $50\text{kg}$, vertical distance in case (a) is $2\text{m}$, mass of sack in case (b) is $25\text{kg}$, vertical distance in case (b) is $4\text{m}$.

Consider the time taken to lift both the sacks is same that is $t$.

Write the expression for the force due to lifting.

$F=mg$

Here,

$m$ is the mass

$g$ is the acceleration due to gravity

Write the expression for work done.

$\begin{array}{l}W=Fd\\ W=mg\times \mathrm{d......\; (I)}\end{array}$

Here,

$d$ is the distance covered

Substitute $50\text{kg}$ for $m$, $2\text{m}$ for $d$, and $10\text{m}/{\text{s}}^2$ for $g$ in equation (I) to get $W$.

$\begin{array}{c}W=\left(50\text{kg}\right)\times \left(10\text{m}/{\text{s}}^2\right)\times 2\text{m}\\ W=1000\text{kg×m}/{\text{s}}^{\text{2}}\cdot \text{m}\times \frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ W=1000\text{N}\cdot \text{m}\end{array}$

Thus, work required in case (a) is $1000\text{N}\cdot \text{m}$.

Substitute $25\text{kg}$ for $m$, $4\text{m}$ for $d$, and $10\text{m}/{\text{s}}^2$ for $g$ in equation (I) to get $W$.

$\begin{array}{c}W=\left(25\text{kg}\right)\times \left(10\text{m}/{\text{s}}^2\right)\times 4\text{m}\\ W=1000\text{kg×m}/{\text{s}}^{\text{2}}\cdot \text{m}\times \frac{1\text{N}}{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ W=1000\text{N}\cdot \text{m}\end{array}$

Thus, work required in case (b) is $1000\text{N}\cdot \text{m}$.

Write the expression for the power.

$P=\frac{W}{t}$...... (II)

Here,

$W$ is the work done

$t$ is the time taken

Substitute $1000\text{N}\cdot \text{m}$ for $W$ and $t$ for $t$ in the equation (II) to get $P$.

$P=\frac{1000\text{N}\cdot \text{m}}{t}$

Thus, the power required in case (a) is $\frac{1000\text{N}\cdot \text{m}}{t}$.

Substitute $1000\text{N}\cdot \text{m}$ for $W$ and $t$ for $t$ in the equation (II) to get $P$.

$P=\frac{1000\text{N}\cdot \text{m}}{t}$

Thus, the power required in case (b) is $\frac{1000\text{N}\cdot \text{m}}{t}$.

The same amount of power is required to lift both the sacks their respective height in same time.

Now, the lighter sack (case b) is lifted about the same distance in half of the time, so the time taken will be $\frac{t}{2}$.

Substitute $1000\text{N}\cdot \text{m}$ for $W$ and $\frac{t}{2}$ for $t$ in the equation (ii) to get $P$.

$\begin{array}{c}P=\frac{1000\text{N}\cdot \text{m}}{\frac{t}{2}}\\ P=2\left(\frac{1000\text{N}\cdot \text{m}}{t}\right)\end{array}$

Thus, the power required to lift in half of the time is twice the power required to lift in time $t$.

Conclusion:

Therefore, the power required to lift both the sacks is same and the power required to lift the lighter sack in half the time is twice to the time required to lift the sack in time $t$.