Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $10.0 m$ : (a) the initially stationary spelunker is accelerated to a speed of $5.00 m / s$ ; (b) he is then lifted at the constant speed of $5.00 m / s$ ; (c) finally he is decelerated to zero speed. How much work is done on the $80.0 k g$ rescuee by the force lifting him during each stage?

Work done on therescuee by force lifting him when,

Accelerated to a speed of $5.00 m/s$ is $8.84 \times 10^{3} J$ .
Lifted with a constant speed of $5.00 m/s$ is $7.84 \times 10^{3} J$ .
Decelerated to zero speed . $6.84 \times 10^{3} J$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The mass of the rescuee is, $m = 80 kg$
Height is, $h = 10.0 m$
Velocity is $v = 5.0 m/s$

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

$P E = m g h$

2.Change in the kineticenergy,

$Δ K E = \frac{1}{2} m v^{2}$

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

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,

$P E = m g h$ (1)

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

$\begin{array}{rcl} P E & = & (80 kg) (9.8 {m/s}^{2}) (10 m) \\ = & 7840 \cdot (1 kg \times 1 {m/s}^{2} \times 1 m \times \frac{1 J}{1 kg \cdot m^{2} {/s}^{2}}) \\ P E & = & 7840 J \end{array}$ (2)

Change Kinetic energy can be calculated as, $Δ K E = \frac{1}{2} m v^{2}$ (3)

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

$Δ K E = \frac{1}{2} (80 kg) {(5 m/s)}^{2} = 1000 \cdot (1 kg \times 1 m^{2} {/s}^{2} \times \frac{1 J}{1 kg \cdot m^{2} {/s}^{2}}) = 1000 J$

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

$W_{1} = Δ K E + Δ P E$

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

$\begin{array}{rcl} W_{1} & = & 1000 J + 7840 J \\ W_{1} & = & 8840 J \\ = & 8.84 \times 10^{3} J \end{array}$

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

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,

$P E = 7840 J$

Thework done can be written as,

$\begin{array}{rcl} W_{2} & = & P E \\ = & 7840 J \\ = & 7.84 \times 10^{3} J \end{array}$

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

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,

$\begin{array}{rcl} Δ K E & = & 0 - \frac{1}{2} (80 kg) {(5 m/s)}^{2} \\ = & - 1000 \cdot (1 kg \times 1 m^{2} {/s}^{2} \times \frac{1 J}{1 kg \cdot m^{2} {/s}^{2}}) \\ = & - 1000 J \end{array}$

From equation (2), potential energy is,

$P E = 7840 J$

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

$W_{3} = Δ K E + Δ P E$

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

$\begin{array}{rcl} W_{3} & = & 7840 J - 1000 J \\ W_{3} & = & 6840 J \\ = & 6.84 \times 10^{3} J \end{array}$

Thus, work done when he is decelerated to zero speed $6.84 \times 10^{3} J$ .

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

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

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

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Understanding the concept

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

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

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

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