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Chapter 9: Problem 46

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 56.61 J.

Step by step solution

01

Understanding Initial and Final Positions

Initially, the ladder is leaning against the wall at a 53掳 angle with the floor. Finally, the ladder is vertical. We need to find the change in height of the ladder's center of mass from its initial to final position to determine the increase in gravitational potential energy.
02

Calculate Initial Height of the Center of Mass

The ladder is uniform, so its center of mass is at its midpoint. Initially, the ladder forms a right triangle with the floor and the wall. The vertical (height) component can be found as: \( h_i = \frac{L}{2} \sin(\theta) \), where \( L = 2.00 \) m is the ladder's length and \( \theta = 53.0^\circ \).
03

Calculate Final Height of the Center of Mass

When the ladder is vertical, its center of mass is at half its height above the ground. Thus, \( h_f = \frac{L}{2} = 1.00 \) m, since the ladder is 2.00 m long and vertical.
04

Calculate Change in Height of the Center of Mass

The change in height \( \Delta h \) is the final height minus the initial height: \( \Delta h = h_f - h_i = 1.00 \, \text{m} - \left( \frac{2.00 \, \text{m}}{2} \sin(53.0^\circ) \right) \).
05

Calculate Change in Gravitational Potential Energy

Gravitational potential energy change is given by \( \Delta U = m g \Delta h \), where \( m = 9.00 \, \text{kg} \) is the mass of the ladder and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Substitute the values to find \( \Delta U \).

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Key Concepts

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

Center of Mass
The center of mass is a fundamental concept in physics to describe the point in a body or system where the entire mass is considered to be concentrated. For a uniform object, which has an even mass distribution, the center of mass is located at the geometrical center.
In the case of a ladder, which is considered uniform, the center of mass will be exactly halfway up the length of the ladder. This point is crucial when analyzing movements and balancing forces.
  • It helps in calculating gravitational potential energy changes.
  • It indicates how the weight of the ladder distributes across its length.
  • Important for determining stability when the ladder is inclined at an angle or moved to a vertical position.
Understanding the center of mass is key in predicting how the ladder will behave when forces are applied, such as when it's repositioned by a worker.
Uniform Ladder
A uniform ladder means that its weight is distributed evenly across its entire length, making calculations and predictions about its behavior simpler. Knowing that the ladder is uniform allows us to easily find the center of mass and make assumptions about how it will react to different forces.
In practice, this uniformity allows you to calculate the potential energy changes by simply focusing on the center of mass, without worrying about the ladder's complex weight distribution.
  • Ensures predictable motion and equilibrium.
  • Allows straightforward calculation of gravitational effects.
  • When calculating angles or potential energy, uniformity simplifies the math given that forces are evenly spread.
Uniform ladders are ideal for problems involving energy, force, or balance due to their dependable properties.
Angle of Inclination
The angle of inclination is the angle the ladder makes with the floor as it leans against the wall. This angle is crucial because it affects both the height at which the center of mass is positioned and the gravitational potential energy.
For a ladder leaning at an angle, the center of mass rises with the sine of the angle. In the problem, the ladder starts at an angle of 53掳 with the floor, which means part of the gravitational force acts vertically.
  • Affects the initial height calculations for gravitational potential energy.
  • Determines how much of the ladder's weight is directed downward compared to along the floor.
  • By changing this angle to vertical, it maximizes the center of mass' height, impacting energy changes.
Understanding this angle helps in predicting how the ladder's center of mass will shift and allows calculating the changes in potential energy accurately.
Vertical Position
When the ladder is moved into a vertical position, its center of mass reaches its maximum possible height above the ground. This is because the full length of the ladder is now aligned perpendicularly to the floor.
In this position, the gravitational potential energy is also maximized. Knowing this allows one to calculate the change in potential energy from the initial leaned position.
  • The center of mass is now at exactly half the ladder's length.
  • This vertical position is stable in terms of having the maximum potential energy.
  • It's crucial for the potential energy difference calculation, since it signifies the highest point the ladder's center of mass reaches.
Recognizing the implications of moving to a vertical position helps in understanding energy changes and optimizing stability and safety when dealing with ladders in practical scenarios.

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Most popular questions from this chapter

A uniform disk has radius \(R$$_0\) and mass \(M$$_0\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \\( \frac{1}{2} \\)\(M_0R_0^2\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_0\), what should be the radius of the circular piece that you remove?

A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

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 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 rad/s\(^2\). What is the kinetic energy of the shell after it has turned through 6.00 rev?
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