/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A helicopter lifts a \(72 \mathr... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 17

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

Short Answer

Expert verified
(a) 11642.4 J; (b) -10584 J; (c) 1058.4 J; (d) 5.42 m/s.

Step by step solution

01

Calculate Gravitational Force

The gravitational force acting on the astronaut can be calculated using the equation: \( F_g = m \cdot g \), where \( m = 72 \mathrm{~kg} \) and \( g = 9.8 \mathrm{~m/s^2} \). Substitute in the values to get: \( F_g = 72 \cdot 9.8 = 705.6 \mathrm{~N} \).
02

Determine the Acceleration Force

The additional acceleration force due to the helicopter can be found using: \( F_a = m \cdot a \), where \( a = \frac{g}{10} = 0.98 \mathrm{~m/s^2} \). Substitute the mass and acceleration to find: \( F_a = 72 \cdot 0.98 = 70.56 \mathrm{~N} \).
03

Calculate Total Lifting Force

The total force exerted by the helicopter (\( F_{h} \)) is the sum of gravitational force and the acceleration force: \( F_{h} = F_g + F_a = 705.6 + 70.56 = 776.16 \mathrm{~N} \).
04

Work Done by Helicopter

Work done by the force from the helicopter can be calculated as: \( W_h = F_{h} \cdot d \), where \( d = 15 \mathrm{~m} \). Thus, \( W_h = 776.16 \cdot 15 = 11642.4 \mathrm{~J} \).
05

Work Done by Gravitational Force

The work done by gravitational force is given by: \( W_g = F_g \cdot d \), with \( F_g = 705.6 \mathrm{~N} \) and \( d = 15 \mathrm{~m} \). Therefore, \( W_g = 705.6 \cdot 15 = 10584 \mathrm{~J} \). Note this work is negative as gravity opposes the upward motion.
06

Calculate Kinetic Energy

The net work done on the astronaut provides her final kinetic energy: \( KE = W_h - |W_g| = 11642.4 - 10584 = 1058.4 \mathrm{~J} \).
07

Determine Speed

The kinetic energy is related to speed by: \( KE = \frac{1}{2} m v^2 \). Solving for \( v \) gives: \( v = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 1058.4}{72}} = \sqrt{29.4} \approx 5.42 \mathrm{~m/s} \).

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion are fundamental in understanding how objects move and interact with forces. These laws describe the relationship between an object, the forces acting upon it, and its motion.

There are three laws:
  • First Law (Inertia): An object will remain at rest or move at a constant speed unless acted upon by an external force. This means that unless the helicopter exerts a force, the astronaut would remain stationary.
  • Second Law (Force and Acceleration): The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This law is best expressed with the equation: \( F = m \cdot a \). In the exercise, the helicopter needs to exert a force (\(776.16 \mathrm{~N}\)) to move the astronaut upwards with the specified acceleration.
  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. When the helicopter pulls the astronaut upwards, the astronaut exerts an equal force downwards on the air by the cable.
Understanding these laws helps visualize how the forces from both the helicopter and gravity act on the astronaut, dictating her motion.
Kinetic Energy
Kinetic energy is a type of energy that an object possesses due to its motion. It is a crucial concept in physics and can be determined with the formula: \( KE = \frac{1}{2} m v^2 \). Here, \( m \) stands for mass, and \( v \) represents velocity.

In our exercise, as the helicopter lifts the astronaut, her velocity increases along with her kinetic energy. Initially, her kinetic energy is zero because she starts from rest. But as work is done on her by the helicopter, she gains kinetic energy. This is calculated based on the work-energy principle, which connects work done and the change in kinetic energy.Kinetic energy gives a measure of how much energy an object has due to movement. It's important in determining not just how fast something is moving but also how much force is needed to stop it. In the exercise, the astronaut's kinetic energy at the end of her upward journey is \(1058.4 \mathrm{~J}\) which indicates the energy she possesses due to her speed of \(5.42 \mathrm{~m/s}\).
Gravitational Force
Gravitational force is one of the four fundamental forces of nature, and it governs the attraction between objects with mass. On Earth, this force gives weight to physical objects and causes them to fall when dropped.

Gravitational force is calculated using the formula: \( F_g = m \cdot g \), where \( m \) is mass and \( g \) is the acceleration due to Earth's gravity, approximately \(9.8 \mathrm{~m/s^2}\). This force always acts downward towards the center of the Earth. In the exercise, the astronaut's gravitational force is \(705.6 \mathrm{~N}\).Gravitational force is integral to understanding free-fall motion and how objects are influenced by gravity. When the helicopter lifts the astronaut, it must exert an upward force greater than this gravitational force to overcome it. The work done by gravity on the astronaut, calculated as a negative value (\(-10584 \mathrm{~J}\)), reflects how gravity opposes the upward motion by attempting to pull her back down to the sea.

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Most popular questions from this chapter

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .\) (a) How much work does the force do in \(10 \mathrm{~min} ?\) (b) What is the average power (in horsepower) of the force?

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

The only force acting on a \(2.0 \mathrm{~kg}\) body as it moves along a positive \(x\) axis has an \(x\) component \(F_{x}=-6 x \mathrm{~N}\), with \(x\) in meters. The velocity at \(x=3.0 \mathrm{~m}\) is \(8.0 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the body at \(x=4.0 \mathrm{~m} ?\) (b) At what positive value of \(x\) will the body have a velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) ?

To pull a \(50 \mathrm{~kg}\) crate across a horizontal frictionless floor, a worker applies a force of \(210 \mathrm{~N}\), directed \(20^{\circ}\) above the horizontal. As the crate moves \(3.0 \mathrm{~m}\), what work is done on the crate by (a) the worker's force, (b) the gravitational force, and (c) the normal force? (d) What is the total work?

A CD case slides along a floor in the positive direction of an \(x\) axis while an applied force \(\vec{F}_{a}\) acts on the case. The force is directed along the \(x\) axis and has the \(x\) component \(F_{a x}=9 x-3 x^{2}\), with \(x\) in meters and \(F_{a x}\) in newtons. The case starts at rest at the position \(x=0\), and it moves until it is again at rest. (a) Plot the work \(\vec{F}_{d}\) does on the case as a function of \(x\). (b) At what position is the work maximum, and (c) what is that maximum value? (d) At what position has the work decreased to zero? (e) At what position is the case again at rest?
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