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A uniform 3.00 -kg rope 24.0 \(\mathrm{m}\) long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to help his partner climb up the cliff. What was the change in potential energy of the rope during this maneuver?

Short Answer

Expert verified
The change in potential energy is -88.29 J.

Step by step solution

01

– Determine the Length of Rope Lowered

The climber lets down half of the 24.0 m rope. Thus, the length of the rope that is lowered is:\[ \text{Length lowered} = \frac{24.0\, \text{m}}{2} = 12.0\, \text{m} \]
02

– Find the Mass of the Lowered Rope

Since the rope is uniform, the mass of the lowered portion can be determined by applying the same proportion as the length:\[ \text{Mass of lowered rope} = \frac{3.00\, \text{kg}}{24.0\, \text{m}} \times 12.0\, \text{m} = 1.50\, \text{kg} \]
03

– Calculate the Change in Height of the Center of Mass

The center of mass of the lowered section moves from the top of the cliff to halfway down the 12.0 m. This means it moves:\[ \text{Change in height} = \frac{12.0\, \text{m}}{2} = 6.0\, \text{m} \] downward.
04

– Calculate the Change in Potential Energy

The change in potential energy is given by the formula:\[ \Delta U = m \cdot g \cdot \Delta h \]Here, \( m = 1.50\, \text{kg} \), \( g = 9.81\, \text{m/s}^2 \), and \( \Delta h = 6.0\, \text{m} \). Substitute these values:\[ \Delta U = 1.50 \times 9.81 \times 6.0 = 88.29 \text{ J} \]Since the rope is lowered, the potential energy decreases by \( 88.29 \text{ J} \).

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Key Concepts

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

Center of Mass
The concept of the center of mass is key when studying the motion and energy of an object like a rope. For a uniform object, such as a homogeneous rope, the center of mass is found at the midpoint of its length. In this exercise, when the climber lowers half of the rope, the center of mass of the lowered section shifts. This halfway point, now at 6 meters below the top of the cliff, becomes crucial to understanding how the system's potential energy changes.
  • The center of mass is the point where an object's mass is evenly distributed.
  • For uniform objects, it lies at the geometric center.
  • In calculations, this point helps determine how energy or force affects the object.
A visual analogy is balancing a yardstick on your finger. If your finger is positioned at the center of mass, the yardstick stays balanced. Similarly, knowing the center of mass of the rope helps calculate how far it descends when half of it is lowered. Grasping this idea can simplify many physics problems, especially those dealing with energy and dynamics. In our problem, it helps us pinpoint the average height change, enabling us to assess the change in potential energy.
Gravitational Potential Energy
Gravitational potential energy refers to the energy an object has because of its position relative to the Earth's surface. The higher the object is, the more potential energy it possesses. In our exercise, as half the rope is lowered, its potential energy at the cliff's edge decreases.
  • Gravitational potential energy ( \( U \)) is calculated using: \( U = m \cdot g \cdot h \)
  • Where \( m \) is mass, \( g \) is the gravitational acceleration (9.81 m/s2), and \( h \) is the height.
  • The change in potential energy ( \( \Delta U \)) is determined by how far and how much mass is moved vertically.
To find the change in potential energy, we need to understand how much mass moves and over what distance, which ties closely to the center of mass concept. Here, as the rope's center of mass descends by 6 meters, the potential energy reduces by 88.29 J, the energy released as the rope is let down.
Mechanics
Mechanics is the branch of physics that deals with motion, forces, and energy. The movement of objects, including how they interact with these forces, forms the crux of problems like the one involving the rope.
  • Mechanics is divided into
  • kinematics (motion description without force) and
  • dynamics (effects of forces on motion).
  • Forces such as gravity influence how objects move and how forces interact over time.
Understanding mechanics means appreciating how all forces, including gravitational, act upon objects like a rope being lowered. In our context, the gravitational force causes the potential energy shift as mass is repositioned. Essentially, mechanics explores the explanation of movement: noting the energy changes gives insight into not just what happens, but why and how it occurs. Mastering mechanics involves comprehensively recognizing such energy transformations and forces at play. In this scenario, energy conservation and transformation underscore why the rope's potential energy decreases as it moves due to gravitational forces.

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Most popular questions from this chapter

What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) long on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An arc 14.0 \(\mathrm{cm}\) long on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

A thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

Energy is to be stored in a 70.0 -kg flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{m.}\) To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the maximum kinetic energy that can be stored in the flywheel?

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

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 0.500 \(\mathrm{kg}\) and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes: (a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bar through one of the balls; (c) an axis parallel to the bar through both balls; (d) an axallel to the bar and 0.500 m from it.

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