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The concepts in this problem are similar to those in Multiple-Concept Example \(4,\) except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 79 -kg person straight up by means of a cable. The person has an upward acceleration of \(0.70 \mathrm{~m} / \mathrm{s}^{2}\) and is lifted from rest through a distance of \(11 \mathrm{~m}\). (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work- energy theorem and find the final speed of the person.

Short Answer

Expert verified
(a) 820.5 N; (b) 9025.5 J; (c) -8532.2 J; (d) 4.4 m/s.

Step by step solution

01

Identify the Forces

The person is subject to two forces: the gravitational force (weight) acting downward and the tension in the cable acting upward. Using Newton's second law, the net force can be expressed as: \( T - mg = ma \).
02

Calculate the Tension in the Cable

Rearrange the equation from step 1 to solve for tension: \( T = m(g + a) \). Substituting the values \( m = 79 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( a = 0.7 \text{ m/s}^2 \), we get: \( T = 79(9.8 + 0.7) = 820.5 \text{ N} \).
03

Find the Work Done by Tension

Work done by the tension is calculated using the formula: \( W_T = T \times d \). Here, \( d = 11 \text{ m} \), so \( W_T = 820.5 \times 11 = 9025.5 \text{ J} \).
04

Calculate the Work Done by Gravitational Force

The work done by gravity is calculated as: \( W_g = -mgd \). Substituting the values: \( W_g = -(79)(9.8)(11) = -8532.2 \text{ J} \).
05

Apply the Work-Energy Theorem to Find Final Speed

According to the work-energy theorem, total work done \( W_{net} = W_T + W_g \) is equal to the change in kinetic energy, \( \ \frac{1}{2}mv^2 - 0 \). Solving for \( v \), we get \( v^2 = \frac{2(W_T + W_g)}{m} \), substituting \( W_T = 9025.5 \text{ J} \) and \( W_g = -8532.2 \text{ J} \), \( v^2 = \frac{2(9025.5 - 8532.2)}{79} \). Calculate \( v \approx 4.4 \text{ m/s} \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental concept used to describe the relationship between an object's mass, its acceleration, and the net force acting upon it. This law is encapsulated in the formula:
  • \( F = ma \) — where \( F \) is the net force applied to the object, \( m \) is the mass, and \( a \) is the acceleration.
When dealing with the tension in a cable lifting a person, we must account for both the gravitational force (weight) and the tension force exerted upward by the cable. These forces impact the person's vertical acceleration. According to Newton's Second Law, the net force is the difference between the upward tension and the downward gravitational force:
  • \( T - mg = ma \) — solves for the net force, where \( T \) is the tension force, \( mg \) is the weight, and \( a \) is the acceleration provided by the tension force.
Rearranging the formula allows us to solve for \( T \), providing the relation: \( T = m(g + a) \). This highlights how the net effect of these forces results in the upward acceleration of the person.
Tension in Cable
In scenarios where objects are being lifted by a cable, such as in a rescue operation, tension plays a crucial role. Tension is the force exerted along the cable, counteracting the gravitational pull.
In this exercise, we need to compute the tension that is required to lift a person with an upward acceleration. We start by considering the forces acting on the individual:
  • Upward force: Tension \( T \)
  • Downward force: Gravitational force \( mg \)
  • Net force resulting in motion: \( ma \) (due to acceleration)
The tension in the cable must not only support the weight of the person but also provide the additional force required to accelerate the person upwards. From Newton's Second Law, the equation \( T = m(g + a) \) shows how tension depends on both gravitational force and the desired acceleration. By substituting the known values of mass, gravity, and acceleration, the tension can be calculated accurately.
Gravitational Work
Gravitational work is a measure of the work done by the gravitational force when an object moves vertically against it. It is crucial in understanding energy transfers in lifting scenarios.
When an object is raised against gravity, gravitational work is negative because the force of gravity opposes the direction of movement. The formula for calculating gravitational work is:
  • \( W_g = -mgd \)
Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( d \) is the height through which the object is lifted. Substituting these values into the formula gives us the precise amount of gravitational work performed during the lift.
In the given exercise, gravitational work is an essential part of calculating the net work done and understanding how it opposes the applied force during the lifting process.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. When analyzing the movement of an object, such as a person being lifted by a helicopter, kinetic energy is influenced by how forces act over distance to change the object's velocity.
The formula for kinetic energy is:
  • \( rac{1}{2}mv^2 \) — where \( m \) is the mass and \( v \) is the velocity.
According to the work-energy theorem, the work done by all the forces acting on an object results in a change in its kinetic energy. This theorem is expressed as:
  • Total work done \( W_{net} = rac{1}{2}mv^2 - 0 \)
In the exercise, both the work done by tension and gravitational work contribute to the net work. After finding these values, you can use them to calculate how much kinetic energy has increased and, consequently, determine the final speed of the person after being lifted.

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Most popular questions from this chapter

Multiple-Concept Example 4 and Interactive LearningWare 6.2 at review the concepts that are important in this problem. \(\mathrm{A} 5.0 \times 10^{4}-\mathrm{kg}\) space probe is traveling at a speed of \(11000 \mathrm{~m} / \mathrm{s}\) through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of \(4.0 \times 10^{5} \mathrm{~N}\) over a distance of \(2500 \mathrm{~km}\). What is the final speed of the probe?

"Rocket man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and is propelled straight upward. At a height of \(16 \mathrm{~m}\), his speed is \(5.0 \mathrm{~m} / \mathrm{s} .\) His mass, including the propulsion unit, has the approximately constant value of \(136 \mathrm{~kg} .\) Find the work done by the force generated by the propulsion unit.

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A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the \(+x\) axis. As a result, the kinetic energy of the sled increases by \(38 \% .\) By what percentage would the sled's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

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