/*! 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 26 An object of mass \(m\) slides d... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 26

An object of mass \(m\) slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is \(h\). The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is (1) \(\mathrm{mgh}\) (2) \(-m g h\) (3) \(-2 m g h\) (4) \(2 \mathrm{mgh}\)

Short Answer

Expert verified
The work to return the object to its initial position is \(2mgh\).

Step by step solution

01

Understanding the Problem

We need to determine the work required to return an object to its initial position after it has descended a height \(h\) and stopped due to friction. The object then needs to be moved back along the same path.
02

Initial Energy Consideration

Initially, the object has a gravitational potential energy of \(mgh\) at the top of the hill due to its height \(h\). As it descends, this energy is converted into kinetic energy and eventually dissipated by friction as it moves along the horizontal path.
03

Energy Dissipation and Work of Friction

As the object moves down and then horizontally along different segments of the path, all initial potential energy \(mgh\) is lost to overcome friction. By the time the object stops, all this energy is consumed by friction forces.
04

Work to Return the Object

To bring the object back to its initial position, the same amount of energy that was dissipated by friction needs to be done in the form of work. This means the force must do work equal to the energy lost, which is \(mgh\) upward against gravity.
05

Total Work Consideration

Since the force has to overcome gravity and retrace the friction path exactly backward, its work is \(mgh\). Therefore, in total, it performs \(2mgh\): \(mgh\) to lift it back and \(mgh\) to overcome friction again.

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Key Concepts

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

Friction
Friction is the resistant force that occurs when two surfaces interact, opposing motion. In our scenario, the object encounters friction while sliding down and during its return journey up the path. Friction varies for different segments of the path but does not depend on how fast or in which direction the object moves. This resistance force is crucial because it converts kinetic energy into heat, causing the object to eventually stop.

When considering friction:
  • It acts opposite to the direction of motion, creating a slowing effect.
  • For each path segment, the friction is constant but varies across different segments.
  • Energy lost due to friction must be overcome to move objects against it.
The total work done against friction will require additional force, especially when the object is to be moved back to its starting point.
Potential Energy
Potential energy represents stored energy due to an object's position, particularly its height in a gravitational field. Initially, when our object is at the top of the hill, it possesses gravitational potential energy measured by the formula:
  • \(PE = mgh\)
where:
  • \(m\) is the mass of the object,
  • \(g\) is the gravitational acceleration, and
  • \(h\) is the height.
As the object descends, this potential energy is converted into kinetic energy as it gains speed. Understanding potential energy helps explain how energy transitions as objects move, particularly in gravitational fields like our hill descent.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As our object slides down the hill, its stored potential energy is transformed into kinetic energy. This transformation can be expressed as:
  • \(KE = \frac{1}{2}mv^2\)
where:
  • \(m\) is mass, and
  • \(v\) is velocity.
The kinetic energy increases as the object accelerates downward, but friction continuously works to decrease this energy, eventually bringing the object to a stop. The interplay between kinetic energy and friction is fundamental, especially as energy dissipates to stop the object, needing replenishment to reverse the path.
Gravitational Force
Gravitational force is the pull experienced by objects due to the Earth's gravity. This force acts downward, affecting both the potential energy at the hill's top and the energy conversion processes during descent and ascent.

Characteristics of gravitational force include:
  • It acts downward, guiding the object's initial movement down the hill.
  • It is responsible for the potential energy the object has at the starting height.
  • During the return journey up the hill, work must counteract this gravitational pull to regain elevation.
While gravity facilitates motion downward by converting potential energy, it poses a challenge when ascending, requiring work to be done against its consistent downward force.

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Most popular questions from this chapter

A man slowly pulls a bucket of water from a well of depth \(h=20 \mathrm{~m}\). The mass of the uniform rope and bucket full of water are \(m=200 \mathrm{~g}\) and \(M 19.9 \mathrm{~kg}\), respectively. Find the work done (in kJ) by the man.

Mark the correct statement(s). (1) Total work done by internal forces of a system on the system is always zero. (2) Total work done by internal forces of a system on the system is sometimes zero. (3) Total work done by internal forces acting between the particles of a rigid body is always zero. (4) Total work done by internal forces acting between the particles of a rigid body is soinetimes zero.

A projectile is launched at angle \(\theta\) to the horizontal from point \(L\) and it hits the target \(T\) on level ground. During the entire motion: $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Magnitude of radial acceleration} & \text { a. } \text{remains constant} \\ \hline \text { ii. } \text{Magnitude of tangential acceleration} & \text { b. } \text{either always decreases or always increases} \\ \hline \text { iii. } \text{Power delivered by gravity} & \text { c. } \text{First increases, then decreases}\\\ \hline \text { iv. } \text{Total acceleration} & \text { d. } \text{First decreases, then increases} \\ \hline \end{array} $$ Now match the given columns and select the correct option from the codes given below. Codes: $$ \begin{array}{lllll} & \text { i. } & \text { ii. } & \text { iii. } & \text { iv. } \\ \text { (1) } &\text { a }& \text { b} & \text { c} & \text { d}\\\ \text { (2) } &\text { c }& \text { d } & \text { c } & \text { a}\\\ \text { (3) } &\text { d }& \text { a } & \text { b } & \text { c }\\\ \text { (4) } &\text { a }& \text { c } & \text { b } & \text { d } \end{array} $$

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F=40\) \(\mathrm{N}\). The kinetic energy of the particle increase \(40 \mathrm{~J}\) in a given interval of time. Then: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) tension in the string is \(40 \mathrm{~N}\) (2) displacement of the block in the given interval of time is \(2 \mathrm{~m}\) (3) work done by gravity is \(-20 \mathrm{~J}\) (4) work done by tension is \(80 \mathrm{~J}\)

One end of a spring of force constant \(k_{1}\) is attached to the ceiling of an elevator. A block of mass \(1.5 \mathrm{~kg}\) is attached to the other end. Another spring of force constant \(k_{2}\) is attached to the bottom of the mass and to the floor of the elevator as shown in the figure. At equilibrium, the deformation in both the springs is equal and is \(40 \mathrm{~cm}\). If the elevator moves with constant acceleration upward, the additional deformation in both the springs is \(8 \mathrm{~cm}\). Find the elevator's acceleration \(\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)\).
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