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Chapter 6: Problem 22

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 \(79-\mathrm{kg}\) 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) Tension is 830.29 N. (b) Work by tension: 9133.19 J. (c) Work by weight: -8524.89 J. (d) Final speed: 3.92 m/s.

Step by step solution

01

Calculate the gravitational force

First, calculate the gravitational force acting on the person using the formula \( F_g = m \cdot g \), where \( m = 79 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Thus, the gravitational force \( F_g = 79 \times 9.81 = 774.99 \text{ N} \).
02

Determine the net force

Using Newton's second law, the net force \( F_{net} \) acting on the person is given by \( F_{net} = m \cdot a \), where \( a = 0.70 \text{ m/s}^2 \). So, \( F_{net} = 79 \times 0.70 = 55.3 \text{ N} \).
03

Calculate the tension in the cable

The tension \( T \) in the cable must overcome the gravitational force and supply the net upward force. Therefore, \( T = F_g + F_{net} = 774.99 + 55.3 = 830.29 \text{ N} \).
04

Work done by the tension in the cable

Calculate the work done by the tension, using the formula \( W = T \cdot d \), where \( d = 11 \text{ m} \). So, the work done \( W = 830.29 \times 11 = 9133.19 \text{ J} \).
05

Work done by the person's weight

The work done by the weight is given by \( W = -F_g \cdot d \) (negative because the force and displacement are in opposite directions). Thus, \( W = -774.99 \times 11 = -8524.89 \text{ J} \).
06

Use work-energy theorem to find final speed

According to the work-energy theorem, the change in kinetic energy is equal to the net work done. The net work done \( W_{net} = 9133.19 - 8524.89 = 608.3 \text{ J} \).\( \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m (0)^2 = 608.3 \). Solving for \( v \): \( \frac{1}{2} \cdot 79 \cdot v^2 = 608.3 \), \( v^2 = \frac{608.3 \times 2}{79} \), \( v^2 = 15.4 \), \( v = \sqrt{15.4} = 3.92 \text{ m/s} \).

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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 fundamental in understanding motion and forces. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. The formula is expressed as \( F = m \cdot a \). This relationship tells us that:
  • The greater the force applied to an object, the greater its acceleration when the mass is constant.
  • The greater the mass of an object, the more force is needed to achieve the same acceleration.
In the exercise, we used Newton's Second Law to determine the net force acting on a person being lifted by a helicopter. The person, with a mass of 79 kg, experiences an upward acceleration of 0.70 m/s虏. Using the formula \( F_{net} = m \cdot a \), we calculated the net force to be 55.3 N. This step is crucial because it helps us understand the system's dynamics and is a precursor for calculating the tension in the lifting cable.
work-energy theorem
The work-energy theorem is a powerful tool in understanding energy transformations in physics. It states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it is written as:\[ W_{net} = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]Where:
  • \( W_{net} \) is the net work done on the object.
  • \( \Delta KE \) is the change in kinetic energy.
  • \( m \) is the mass.
  • \( v_f \) is the final velocity.
  • \( v_i \) is the initial velocity.
In this homework problem, the work-energy theorem was used to determine the final speed of the person after being lifted by the cable. Initially at rest, the person gains kinetic energy due to the net work done by both the tension in the cable and gravity. Calculating this change in kinetic energy allowed us to solve for the final speed, which resulted in approximately 3.92 m/s.
gravitational force calculation
Calculating gravitational force is important when dealing with vertical movements where gravity plays a crucial role. The gravitational force on an object, also known as its weight, can be calculated using:\[ F_g = m \cdot g \]Where:
  • \( F_g \) is the gravitational force.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately 9.81 m/s虏 on Earth's surface.
In the given exercise, the gravitational force acting on the 79-kg person was found to be approximately 774.99 N. Understanding this force is crucial as it opposes the tension in the cable that lifts the person. The balance and interaction between gravitational force and tension directly affect the motion of the person being lifted, dictating the necessary force and work adjustments needed to achieve the desired motion.

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Most popular questions from this chapter

A gymnast is swinging on a high bar. The distance between his waist and the bar is \(1.1 \mathrm{m},\) as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

The drawing shows a skateboarder moving at \(5.4 \mathrm{m} / \mathrm{s}\) along a horizontal section of a track that is slanted upward by \(48^{\circ}\) above the horizontal at its end, which is \(0.40 \mathrm{m}\) above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height \(H\) to which she rises above the end of the track.

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta .\) One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0}\). On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

A \(16-\mathrm{kg}\) sled is being pulled along the horizontal snow-covered ground by a horizontal force of \(24 \mathrm{N}\). Starting from rest, the sled attains a speed of \(2.0 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{m} .\) Find the coefficient of kinetic friction between the runners of the sled and the snow.

A 2.00-kg rock is released from rest at a height of 20.0 m. Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: \(20.0,10.0,\) and \(0 \mathrm{m}\)
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