/*! 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 64 Engineers are designing a system... [FREE SOLUTION] | 91Ó°ÊÓ

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Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. \(\mathbf{P 9 . 6 4}\) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to \(15.0 \mathrm{~kg}\) and allowed to fall through \(5.00 \mathrm{~m},\) it gives \(250.0 \mathrm{~J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0 \mathrm{~kg}\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the \(15.0 \mathrm{~kg}\) mass be moving on Mars just as the drum gained \(250.0 \mathrm{~J}\) of kinetic energy?

Short Answer

Expert verified
a) The mass would have to fall through a distance of approximately 4.48 m on Mars to deliver the same kinetic energy to the drum as is on Earth. b) The speed of the mass on Mars would be approximately 7.75 m / s just as the drum gains 250.0 J of kinetic energy.

Step by step solution

01

Determine the gravitational potential energy on Earth

Use the formula for gravitational potential energy \((PE = mgh)\) where \(m\) is the mass, \(g\) is the acceleration due to gravity and \(h\) is the height. On Earth, \(g = 9.81 m / s^{2} .\) Thus, \(PE = 15.0 kg * 9.81 m / s^{2} * 5.0 m = 735.75 J .\)
02

Compute the kinetic energy of the drum on Earth

Given that the mass gives \(250.0 J\) of kinetic energy to the drum, it can be inferred that \(PE - KE = 735.75 J - 250.0 J = 485.75 J .\) This difference has been used for the work done in rotating the drum, and it's due to the conservation of energy.
03

Find the height that the mass must fall on Mars (Part a)

On Mars, the acceleration due to gravity is \(3.71 m / s^{2} .\) The kinetic energy on Mars must be same as that on Earth, \(250.0 J\), by the conservation of energy. Thus, to compute the height \(h = KE / (m * g) = 250.0 J / (15.0 kg * 3.71 m / s^{2}) = 4.48 m .\) Hence, the mass would have to fall through a distance of about 4.48 m to deliver the same amount of kinetic energy to the drum on Mars as it did on Earth.
04

Calculate the speed of the mass on Mars (Part b)

The kinetic energy follows this formula \(KE = 0.5 mv^{2}\) where \(v\) is the velocity of the mass just before it hits the ground. We can rearrange and solve for \(v\), \(v = sqrt((2 * KE) / m) = sqrt((2 * 250.0 J) / 15.0 kg) = 7.75 m / s .\)

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. It's calculated using the formula:
  • \(PE = mgh\)
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity,
  • \(h\) is the height above the reference point (usually the ground or the floor).
This energy represents the potential for an object to do work as a result of its position. In simpler terms, it’s the energy stored when you lift an object against gravity.
On Earth, for the exercise at hand, the 15kg mass at 5 meters height possesses a potential energy of approximately 735.75 Joules, calculated with Earth's gravity \(9.81 m/s^2\).
When this system is evaluated on Mars, where gravity is weaker at \(3.71 m/s^2\), an object requires less height to gain the same potential energy because each meter contributes less to the energy due to the lower gravitational pull.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. The faster an object moves, the more kinetic energy it has. It's given by the equation:
  • \(KE = \frac{1}{2}mv^2\)
  • \(m\) is the mass,
  • \(v\) is the velocity.
The kinetic energy transformation from gravitational potential energy demonstrates the Conservation of Energy principle, where energy is transferred but the total amount remains constant.
In the given problem, the mass, while descending, transfers its gravitational potential energy into kinetic energy, giving the drum 250 Joules. This conversion process remains the same on both Earth and Mars as energy transformation principles hold regardless of location.
On Mars, when the mass reaches the same amount of kinetic energy, its speed calculates to be 7.75 meters per second, reaffirming the idea that even though the fall differs, the converted kinetic energy remains consistent.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as \(g\), affects how quickly an object accelerates when falling. It varies depending on the celestial body. On Earth, \(g = 9.81 m/s^2\), while on Mars, it's significantly less at \(3.71 m/s^2\). This difference plays a crucial role in problems involving gravitational potential energy and kinetic energy as it dictates how energy transformations occur during free fall.
For our exercise, the variation of gravity between Earth and Mars means that to achieve the same energy arrangements, such as the kinetic energy imparted to the drum, the falling distance must adapt. Mars’ weaker gravity necessitates that objects need less height to achieve the same energy transfer to kinetic energy compared to Earth.
This is central to the system design in the exercise because understanding and calculating these influences allow engineers to create adaptable solutions that work efficiently in diverse gravitational environments.

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Most popular questions from this chapter

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A uniform bar has two small balls glued to its ends. The bar is \(2.00 \mathrm{~m}\) long and has mass \(4.00 \mathrm{~kg},\) while the balls each have mass \(0.300 \mathrm{~kg}\) and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and \(0.500 \mathrm{~m}\) from it.

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A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

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