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A uniform 2.00 -m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of \(53.0^{\circ}\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

Short Answer

Expert verified
The increase in gravitational potential energy is approximately 35.6 J.

Step by step solution

01

Calculate Initial Height

The initial height of the ladder can be found using the vertical component of the ladder's length. Using trigonometry, we calculate it as \( h_i = L \cdot \sin(\theta) \), where \( L = 2.00 \text{ m} \) and \( \theta = 53.0^\circ \). \[ h_i = 2.00 \times \sin(53.0^\circ) \] The sine of 53 degrees is approximately 0.7986, so:\[ h_i = 2.00 \times 0.7986 \approx 1.597 \text{ m} \].
02

Determine Final Height

When the ladder is positioned vertically, its entire length is the height. Therefore, the final height is simply equal to the length of the ladder:\[ h_f = 2.00 \text{ m} \].
03

Calculate Change in Height

The change in height \( \Delta h \) is the difference between the final and initial heights:\[ \Delta h = h_f - h_i = 2.00 \text{ m} - 1.597 \text{ m} = 0.403 \text{ m} \].
04

Calculate Change in Gravitational Potential Energy

The change in gravitational potential energy \( \Delta U \) can be calculated as:\[ \Delta U = m \cdot g \cdot \Delta h \] where \( m = 9.00 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Plug in the values:\[ \Delta U = 9.00 \times 9.81 \times 0.403 \approx 35.6 \text{ J} \].

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Key Concepts

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

Trigonometry in Physics
In physics, trigonometry is a fundamental tool used to solve many problems, especially those involving angles and geometric relationships. When a ladder leans against a wall, trigonometry helps to determine different components like angles and distances. This is crucial for determining height when the ladder forms an angle with the horizontal ground.

For example, the exercise involves a ladder making an angle with the floor. We find the height against the wall using the sine function. Trigonometry gives us these helpful relations:
  • The sine function, written as \( \sin(\theta) \), relates the angle to the opposite side and hypotenuse of a right triangle.
  • Here, \( h_i = L \cdot \sin(\theta) \), where \( h_i \) is the vertical height, \( L \) is the ladder's length, and \( \theta \) is the angle.
By calculating \( \sin(53.0^\circ) \), we find part of the ladder's height above the ground when it is not fully vertical. This makes trigonometry an indispensable part of understanding the physics of ladders.
Ladder Problem
Ladder problems are common in physics exercises and help illustrate the concepts of static equilibrium and different forces at play. When dealing with ladders, it's essential to understand not only the forces but also the angles involved. In this example,
  • An initial angle of \( 53.0^\circ \) defines how the ladder leans against the wall.
  • The ladder's calculations involve assessing both gravitational potential energy and the center of mass.
To solve ladder problems effectively, one must carefully analyze the setup, initial conditions, and physical setups such as walls and floors. Initially, our ladder makes a certain angle with the ground, meaning part of its length contributes only vertically. Lifting the ladder until it stands upright changes this setup.

These principles find practical applications in engineering and construction, showing how forces exert upon structures.
Mechanical Energy Conservation
Mechanical energy conservation involves understanding how energy transfers and transforms in a system. In this scenario, the question focuses on gravitational potential energy, a type of mechanical energy related to an object's height and mass.

Initially, the ladder has a certain gravitational potential energy given by its vertical positioning defined by the height \( h_i \). Using principles of mechanical energy, we calculate the energy change when the ladder changes position:
  • The gravitational potential energy depends on both height and mass, expressed as \( U = m \cdot g \cdot h \), where \( m \) is mass, and \( g \) is Earth's gravitational pull (9.81 m/s²).
  • By shifting the ladder from its original angle to stand vertically, there is a distinct change in potential energy.
This change, calculated as \( \Delta U = m \cdot g \cdot \Delta h \), denotes the energy used to raise the ladder. Such calculations illustrate the principles of energy conservation, where the initial setup transforms, and energy is neither created nor destroyed—only shifted from one form to another.

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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 thin, flat, uniform disk has mass \(M\) and radius \(R .\) A circular hole of radius \(R / 4,\) centered at a point \(R / 2\) from the disk's center, is then punched in the disk. (a) Find the moment of inertia of the disk with the hole about an axis through the original center of the disk, perpendicular to the plane of the disk. (Hint: Find the moment of inertia of the piece punched from the disk.) (b) Find the moment of inertia of the disk with the hole about an axis through the center of the hole, perpendicular to the plane of the disk.

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm}\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

Measuring \(I .\) As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.740 \(\mathrm{m}\) and find that it weighs 280 \(\mathrm{N.}\) You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang an 8.00 -kg mass from the free end of the rope, as shown in Fig. 9.17 . You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 5.00 \(\mathrm{m} / \mathrm{s}\) after it has descended 2.00 \(\mathrm{m} .\) (a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center? (b) Your boss tells you that a larger \(I\) is needed. He asks you to design a wheel of the same mass and radius that has \(I=19.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\) How do you reply?

The Kinetic Energy of Walking. If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \(\frac{1}{2} M V^{2}\).However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\%\) . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ} )\) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 75 -kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

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