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Chapter 4: Problem 87

The alarm at a fire station rings and an \(86-\mathrm{kg}\) fireman, starting from rest, slides down a pole to the floor below (a distance of \(4.0 \mathrm{m}\) ). Just before landing, his speed is \(1.4 \mathrm{m} / \mathrm{s}\). What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Short Answer

Expert verified
The magnitude of the kinetic frictional force is approximately 810 N.

Step by step solution

01

Determine the total mechanical energy change

The fireman starts from rest, so his initial kinetic energy is zero. The initial potential energy is given by \(mgh\), where \(m = 86\, \text{kg}\), \(g = 9.81\, \text{m/s}^2\) is the acceleration due to gravity, and \(h = 4.0\, \text{m}\) is the height. Thus, the initial potential energy is \(86 \times 9.81 \times 4.0\). The final kinetic energy just before he hits the ground is \(\frac{1}{2}mv^2\), where \(v = 1.4\, \text{m/s}\). Substituting the values, the final kinetic energy is \(\frac{1}{2} \times 86 \times 1.4^2\).
02

Calculate the work done by friction

The work-energy principle states that the work done by non-conservative forces (friction) is equal to the change in mechanical energy. The change in mechanical energy is the difference between the initial potential energy and the final kinetic energy. Therefore, \(W_{\text{friction}} = \text{Initial Potential Energy} - \text{Final Kinetic Energy}\).
03

Solve for the frictional force

The work done by friction is also given by \(W_{\text{friction}} = f_k \cdot d\), where \(f_k\) is the kinetic frictional force and \(d = 4.0\, \text{m}\) is the distance slipped. Rearrange this to find \(f_k = \frac{W_{\text{friction}}}{d}\). Substitute the value of \(W_{\text{friction}}\) calculated in Step 2 and \(d = 4.0\, \text{m}\) into the equation to find \(f_k\).

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Key Concepts

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

Mechanical Energy
Mechanical energy is the sum of kinetic energy and potential energy in a system.
In the exercise, the fireman's mechanical energy includes his potential energy at the top of the pole and his kinetic energy as he descends.
The initial mechanical energy is all in the form of potential energy because he starts from rest.
As he slides down, some of this potential energy converts into kinetic energy.
However, not all potential energy becomes kinetic energy because some energy is lost to friction.
  • Initial Potential Energy: This is given by the formula \(PE = mgh\), where \(m\) is mass, \(g\) is the gravitational force, and \(h\) is height.
  • Kinetic Energy: As he starts to move, some potential energy transforms into kinetic energy, calculated using \(KE = \frac{1}{2}mv^2\).
Understanding mechanical energy helps us analyze how energy is conserved or transformed in physical processes.
Work-Energy Principle
The work-energy principle states that the work done by forces on an object equals its change in mechanical energy.
This principle combines concepts of work and energy to describe how external forces change an object's energy.
Using this principle, we can calculate the work done by the frictional force on the fireman as he slides down.
  • Total Work Done: This equals the change in mechanical energy, which in our exercise accounts for the difference between initial potential energy and final kinetic energy.
  • Friction as a Non-Conservative Force: Friction causes some mechanical energy to be lost as heat, accounting for energy not converted into kinetic energy.
This principle is crucial in understanding how forces other than gravity impact energy changes in a system.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion.
For the fireman, kinetic energy only appears right before he reaches the floor as he starts from rest.
The formula for kinetic energy is \(KE = \frac{1}{2} mv^2\). Here, it shows us how movement converts energy directly from potential to kinetic.
  • Dependence on Speed: Kinetic energy is directly proportional to the square of velocity, showing higher speeds result in significantly more energy.
  • Mass Factor: The fireman's mass also plays a role, with heavier objects having more kinetic energy at the same speed.
Kinetic energy aids in understanding the dynamic side of energy transformations.
Potential Energy
Potential energy refers to the energy stored within an object due to its position or configuration.
For the fireman, this energy is due to his position at the top of the pole, related to gravity.
The gravitational potential energy is calculated as \(PE = mgh\).
  • Height's Role: The higher the position, the more potential energy is stored.
  • Gravitational Influence: The constant \(g\) represents Earth's gravitational pull, affecting how potential energy decreases as he falls.
Potential energy gives insight into how stored energy can become active as movement energy.

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Most popular questions from this chapter

Two objects \((45.0\) and \(21.0 \mathrm{kg}\) ) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.

A 292 -kg motorcycle is accelerating up along a ramp that is inclined \(30.0^{\circ}\) above the horizontal. The propulsion force pushing the motorcycle up the ramp is \(3150 \mathrm{N}\), and air resistance produces a force of \(250 \mathrm{N}\) that opposes the motion. Find the magnitude of the motorcycle's acceleration.

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs \(31 \mathrm{N}\). The coefficient of static friction between his hands and the book is \(0.40 .\) To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of \(151 \mathrm{N}\) on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{N} .\) Find the magnitude and direction of the tension that the right section of the wire sustains.

The weight of an object is the same on two different planets. The mass of planet \(A\) is only sixty percent that of planet \(B\). Find the ratio \(r_{A} / r_{B}\) of the radii of the planets.
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