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Expert-verified Found in: Page 172 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of ${\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbit{m}}$: (a) the initially stationary spelunker is accelerated to a speed of ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$; (b) he is then lifted at the constant speed of${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$ ; (c) finally he is decelerated to zero speed. How much work is done on the ${\mathbf{80}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{}}{\mathbf{\text{​}}}{\mathbit{k}}{\mathbit{g}}{\mathbf{}}$rescuee by the force lifting him during each stage?

Work done on therescuee by force lifting him when,

1. Accelerated to a speed of $5.00\text{m/s}$ is $8.84×{10}^{3}\text{J}$.
2. Lifted with a constant speed of $5.00\text{m/s}$is$7.84×{10}^{3}\text{J}$.
3. Decelerated to zero speed .$6.84×{10}^{3}\text{J}$
See the step by step solution

## Step 1: Given data

1. The mass of the rescuee is,$m=80\text{kg}$
2. Height is,$h=10.0\text{m}$
3. Velocity is$v=5.0\text{m/s}$

## Step 2: Understanding the concept

By usingthe concept of change in kinetic as well as potential energy, we can find the work done during each step.

Formula:

1.The potential

$PE=mgh$

2.Change in the kineticenergy,

$\Delta KE=\frac{1}{2}m{v}^{2}$

Here,g is the gravitational acceleration whose value is $9.8{\text{m/s}}^{\text{2}}$

## Step 3: (a) Calculate work done on the 80.0 kg rescuee if initially stationary spelunker is accelerated to a speed of

Potential energy can be calculated as,

$PE=mgh$ (1)

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}PE& =& \left(80\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(10\text{m}\right)\\ & =& 7840·\left(1\text{kg}×1{\text{m/s}}^{\text{2}}×1\text{m}×\frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ PE& =& 7840\text{J}\\ & & \end{array}$ (2)

Change Kinetic energy can be calculated as, $\Delta KE=\frac{1}{2}m{v}^{\text{2}}$ (3)

Substituting the values in the above expression, and we get,

$\Delta KE=\frac{1}{2}\left(80\text{kg}\right){\left(5\text{m/s}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1000·\left(1\text{kg}×1{\text{m}}^{2}{\text{/s}}^{\text{2}}×\frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\phantom{\rule{0ex}{0ex}}=1000\text{J}\phantom{\rule{0ex}{0ex}}$

From the work-energy theorem, the total change in energy and the work done on the system will be equal; then, we can write,

${W}_{1}=\Delta KE+\Delta PE$

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}{W}_{\text{1}}& =& 1000\text{J}+7840\text{J}\\ {W}_{\text{1}}& =& 8840\text{J}\\ & =& 8.84×{10}^{3}\text{J}\\ & & \end{array}$

Thus, work done when spelunker is accelerated to a speed of $5.00\text{m/s}$ is $8.84×{10}^{3}\text{J}$.

## Step 4: (b) Calculate work done on the rescuee if he is lifted at the constant speed of 5.00 m/s

In this case, the speed is constant.

So total energy will be the potential energy, and kinetic energy will remain the same.

From equation (2), the potential energy will be,

$PE=7840\text{J}$

Thework done can be written as,

$\begin{array}{rcl}{W}_{\text{2}}& =& PE\\ & =& 7840\text{J}\\ & =& 7.84×{10}^{3}\text{J}\\ & & \end{array}$

Thus, the work done when he is lifted with a constant speed of $5.00\text{m/s}$is $7.84×{10}^{3}\text{J}$.

## Step 5: (c) Calculate work done on the 80.0  kg rescuee when he is decelerated to zero speed

In this case, the velocity decreases to zero from 5 m/s.

Then from equation 3, the decrease in kinetic energy can be calculated as,

$\begin{array}{rcl}\Delta KE& =& 0-\frac{1}{2}\left(80\text{kg}\right){\left(5\text{m/s}\right)}^{2}\\ & =& -1000·\left(1\text{kg}×1{\text{m}}^{2}{\text{/s}}^{\text{2}}×\frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ & =& -1000\text{J}\\ & & \end{array}$

From equation (2), potential energy is,

$PE=7840\text{J}$

From the work-energy theorem, the total change in energy and the work done on the system will be equal; then, we can write,

${W}_{3}=\Delta KE+\Delta PE$

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}{W}_{3}& =& 7840\text{J}-1000\text{J}\\ {W}_{\text{3}}& =& 6840\text{J}\\ & =& 6.84×{10}^{3}\text{J}\\ & & \end{array}$

Thus, work done when he is decelerated to zero speed $6.84×{10}^{3}\text{J}$. ### Want to see more solutions like these? 