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Chapter 9: Problem 45

Cave Rescue 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 \(10.0 \mathrm{~m}:\) (a) the initially stationary spelunker is accelerated to a speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (b) he is then lifted at the constant speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (c) finally he is slowed to zero speed. How much work is done on the \(80.0 \mathrm{~kg}\) rescuee \(\quad F_{x}\) by the force lifting him during each stage?

Short Answer

Expert verified
7848 J

Step by step solution

01

Identify given information

The mass of the spelunker is 80.0 kg. There are three stages with a vertical distance of 10.0 m each. The initial speed is 0 m/s, the speed during the lift is 5.00 m/s, and the final speed is 0 m/s.
02

Calculate work done in stage (a)

Stage (a) involves accelerating the spelunker from rest to a speed of 5.00 m/s. Use the work-energy theorem: \( W = \frac{1}{2} m v^2 \). Substitute the values: \( W = \frac{1}{2} \times 80.0 \times (5.00)^2 = 1000 \text{ J} \).
03

Calculate work done in stage (b)

In stage (b), the spelunker is lifted at a constant speed (5.00 m/s). The work done is against gravity: \( W = mgh \). Given the spelunker is lifted 10.0 m, substitute the values: \( W = 80.0 \times 9.81 \times 10.0 = 7848 \text{ J} \).
04

Calculate work done in stage (c)

Stage (c) involves decelerating the spelunker from 5.00 m/s to rest. The work done is the negative of the work done in stage (a): \( W = -\frac{1}{2} m v^2 \). Substitute the values: \( W = -1000 \text{ J} \).
05

Sum the total work done

Sum the work done in all three stages: \( W_{\text{total}} = 1000 + 7848 - 1000 = 7848 \text{ J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work-Energy Theorem
The work-energy theorem is a fundamental concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. For example, in the cave rescue scenario, the spelunker's kinetic energy changes as he is lifted.
In stage (a), we calculate the work required to accelerate the spelunker from rest to 5 m/s using the formula:
\(\text{Work} = \frac{1}{2} m v^2\)
Here, \(m\) is the mass and \(v\) is the final velocity. This gives us the work done to change the kinetic energy to that velocity.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. The formula for kinetic energy is:
\(\text{KE} = \frac{1}{2} m v^2\).
In our problem, the spelunker gains kinetic energy in stage (a) as he accelerates to 5 m/s. The work done on him is converted into kinetic energy.
This energy tells us how much work was needed to move the spelunker to his new speed. When he moves at 5 m/s, his kinetic energy is at its maximum for this scenario.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object has due to its height above the ground. It is calculated using: \(\text{GPE} = mgh\),
where \(m\) is the mass, \(g\) is the acceleration due to gravity (9.81 m/s虏), and \(h\) is the height.
In stage (b), as the spelunker is lifted vertically at a constant speed, his GPE increases. The work done against gravity here accounts for this increase, with the formula showing us the energy needed to lift him.
Constant Speed Motion
Constant speed motion is when an object's speed does not change over time. In this condition, the object's kinetic energy remains the same since there is no acceleration.
In stage (b) of the rescue, the spelunker moves at a constant speed of 5 m/s. The work done during this stage goes into overcoming gravitational force, not changing kinetic energy. This stage shows how work is used to maintain motion against external forces like gravity.
Deceleration
Deceleration is the process of reducing the speed of an object. In physics, it's considered as negative acceleration. The kinetic energy of the object decreases, converting energy to other forms, such as heat or sound, depending on the context.
In stage (c), the spelunker is decelerated from 5 m/s to rest. The work needed to do this is the negative of the work done to accelerate him initially. This means energy is taken from the kinetic energy, reducing it to zero as the spelunker comes to a stop.

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Most popular questions from this chapter

Large Meteorite vs. TNT On August 10,1972, a large meteorite skipped across the atmosphere above western United States and Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky (see Fig. \(9-22\) for a similar event). The meteorite's mass was about \(4 \times 10^{6} \mathrm{~kg}\) : its speed was about \(15 \mathrm{~km} / \mathrm{s}\). Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of \(\mathrm{TNT}\), which is \(4.2 \times 10^{15} \mathrm{~J}\). (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

Transporting Boxes Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At a certain location the conveyor belt moves for \(2.0 \mathrm{~m}\) up an incline that makes an angle of \(10^{\circ}\) with the horizontal, then for \(2.0 \mathrm{~m}\) horizontally, and finally for \(2.0 \mathrm{~m}\) down an incline that makes an angle of \(10^{\circ}\) with the horizontal. Assume that a \(2.0 \mathrm{~kg}\) box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box (a) as the box moves up the \(10^{\circ}\) incline, (b) as the box moves horizontally, and (c) as the box moves down the \(10^{\circ}\) incline?

A Particle Moves A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}-(c \mathrm{~m}) \hat{\mathrm{j}}\) while force \(\vec{F}=\) \((2 \mathrm{~N}) \mathrm{i}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

Force on a Particle \(A\) force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}+(c \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle as the particle goes through displacement \(\vec{d}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((2.0 \mathrm{~m}) \hat{\mathrm{j}} .\) (Other forces also act on the particle.) What is the value of \(c\) if the work done on the particle by force \(\vec{F}\) is (a) zero, (b) \(17 \mathrm{~J}\), and (c) \(-18 \mathrm{~J} ?\)

A Locomotive A locomotive with a power capability of \(1.5 \mathrm{MW}\) can accelerate a train from a speed of \(10 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~min}\). (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train as functions of time (in seconds) during the \(6.0\) min interval. (d) Find the distance moved by the train during the interval.
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