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Chapter 9: Problem 52

A uniform \(2.00-\mathrm{m}\) ladder of mass 9.00 \(\mathrm{kg}\) 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. How much work did this person do against gravity?

Short Answer

Expert verified
The work done against gravity is approximately 17.75 J.

Step by step solution

01

Understanding Work Done Against Gravity

The work done against gravity is equal to the change in gravitational potential energy of the ladder. Gravitational potential energy (GPE) is calculated using the formula: \[\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h\]where \(m\) is the mass of the ladder, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(\Delta h\) is the change in height of the center of mass of the ladder.
02

Calculate Original Height of Center of Mass

Initially, the ladder's center of mass is halfway along its length at an angle of \(53^{\circ}\) to the ground. The initial height \(h_i\) of the center of mass is given by:\[h_i = \frac{L}{2} \cdot \sin(53^{\circ})\]where \(L = 2.00 \, \mathrm{m}\). So,\[h_i = 1.00 \, \mathrm{m} \cdot \sin(53^{\circ}) \approx 1.00 \, \mathrm{m} \cdot 0.7986 = 0.7986 \, \mathrm{m}\]
03

Calculate Final Height of Center of Mass

When the ladder is vertical, the center of mass is directly above the ground at half its length. Thus, the final height \(h_f\) of the center of mass is:\[h_f = \frac{L}{2} = 1.00 \, \text{m}\]
04

Calculate Change in Height of the Center of Mass

The change in height \(\Delta h\) of the ladder's center of mass is the difference between the final and initial heights:\[\Delta h = h_f - h_i = 1.00 \, \text{m} - 0.7986 \, \text{m} = 0.2014 \, \text{m}\]
05

Calculate Work Done Against Gravity

Now, we calculate the work done against gravity using the change in gravitational potential energy:\[\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h = 9.00 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.2014 \, \text{m} \approx 17.75 \, \text{J}\]

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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 a form of energy associated with the position of an object in a gravitational field. It's determined by the object's mass, the height of its center of mass relative to a reference point, and the strength of the gravitational field. This energy can be calculated using the formula:
  • \[\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h\]
  • where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) on Earth), and \(\Delta h\) is the change in height of the object's center of mass.
This concept is crucial when evaluating work done against gravity. For instance, when lifting an object, you increase its gravitational potential energy. Similarly, if an object moves upwards, its gravitational potential energy increases. Understanding this energy conversion helps explain why work must be done to lift objects and how this work is quantified.
Center of Mass
The center of mass of an object is a point where the mass is evenly distributed in all directions. It's essentially the balance point of that object. For uniform objects like a ladder, the center of mass is found at the midpoint if the object has equal density throughout.
  • When you know its position, it helps in calculating the potential energy change as you tilt or move the object.
  • For the ladder leaning against a wall, the center of mass is initially at half the ladder's height.
As you alter the ladder's position, such as moving it to a vertical stance, the vertical position of the center of mass changes. This change is crucial to calculate how much work is done as you are working against the force of gravity. Essentially, knowing the center of mass allows us to compute how far it is displaced vertically, which forms a key component in energy calculations.
Angle of Inclination
The angle of inclination refers to the angle formed between the base of an object and the horizontal ground. This angle plays a significant role in determining the height of the center of mass when dealing with inclined objects like a ladder.
  • If you have an angle, you can determine the object's height above the ground using trigonometric functions, particularly sine.
  • For a ladder leaning against a wall, the height of the center of mass when inclined is calculated with:

  • Height Formula for Inclined Objects

    \[h_i = \frac{L}{2} \cdot \sin(\text{angle of inclination})\]
  • Using the sine function to find the vertical component based on the angle of inclination.
Adjusting the angle of inclination alters the height of the center of mass, affecting the gravitational potential energy and, consequently, the work done against gravity when moving the object. Understanding these principles helps in computing the slight variations in gravitational potential energy encountered when changing the angle of inclination.

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Most popular questions from this chapter

A high-speed flywheel in a motor is spinning at 500 \(\mathrm{rpm}\) when a power failure suddenly occurs. The flywheel has mass 40.0 \(\mathrm{kg}\) and diameter 75.0 \(\mathrm{cm}\) . The power is off for 30.0 \(\mathrm{s}\) , and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s}\) . What angular velocity is required to achieve these values?

How I Scales. If we multiply all the design dimensions of an object by a scaling factor \(f\) , its volume and mass will be multiplied by \(f^{3}\) . (a) By what factor will its moment of inertia be multiplied? (b) If a \(\frac{1}{48}\) - scale model has a rotational kinetic energy of \(2.5 \mathrm{J},\) what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 \(\mathrm{kg}\) and radius 20.0 \(\mathrm{cm}\) . A 1.50 \(\mathrm{kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. 9.32\()\) , and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 \(\mathrm{J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

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.
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