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Chapter 8: Problem 128

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

Short Answer

Expert verified
(a) 9.2 m/s (b) 4.75 m/s

Step by step solution

01

Identify the Known Quantities

We have the following information: mass of firefighter, \( m = 70 \, \text{kg} \), distance down the pole, \( h = 4.3 \, \text{m} \), gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \).
02

Calculate Speed with No Friction (Part a)

We'll use the conservation of energy: the potential energy at the top is converted into kinetic energy at the bottom. Initial potential energy \( U_i = mgh \), and final kinetic energy \( K_f = \frac{1}{2}mv^2 \). Setting \( U_i = K_f \) gives: \[ mgh = \frac{1}{2}mv^2 \] Simplifying and solving for \( v \): \[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 4.3 \, \text{m}} \approx 9.2 \, \text{m/s} \]
03

Calculate Speed with Friction (Part b)

When friction is involved, the work done by friction is subtracted from the potential energy: \( mgh - f_s h = \frac{1}{2}mv^2 \). Friction force \( f_s = 500 \, \text{N} \). Substitute and solve for \( v \): \[ 70 \times 9.8 \times 4.3 - 500 \times 4.3 = \frac{1}{2} \times 70 \times v^2 \]\[ 2940 - 2150 = 35v^2 \] \[ 790 = 35v^2 \] \[ v^2 = \frac{790}{35} \approx 22.57 \] \[ v \approx \sqrt{22.57} \approx 4.75 \, \text{m/s} \]

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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 represents the energy an object holds due to its position in a gravitational field. When an object is at a height, it has the potential to do work by falling. For the firefighter sliding down the pole, her initial potential energy can be calculated using the formula \( U_i = mgh \). Here, \( m \) is the mass of the firefighter, \( g \) is the acceleration due to gravity, and \( h \) is the height from which she starts. In this problem, the firefighter's mass is 70 kg, the height is 4.3 meters, and gravity is 9.8 m/s虏. To find the gravitational potential energy, multiply these values. Initially, this energy is at its maximum since she hasn't moved downward yet, and this energy will be transformed as she slides.
Kinetic Energy
Kinetic energy is the energy of motion. It's determined by how fast an object is moving and its mass. The formula to calculate kinetic energy is \( K = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. As the firefighter begins to slide down the pole, her potential energy is converted to kinetic energy because she gains speed.
For part (a) of the exercise, with negligible friction, all of the gravitational potential energy is converted into kinetic energy as she reaches the ground. By equating the potential and kinetic energy, we derive her speed. For part (b), where friction is present, the kinetic energy will be lower because some energy is lost to friction. This affects her final speed as calculated using the work-energy principle.
Work-Energy Principle
The work-energy principle ties together work done on a system and its energy changes. It states that the work done by all forces acting on an object equals the change in kinetic energy. In the firefighter's scenario, if she holds the pole firmly, friction does work against her motion.
The friction force, \( f_s = 500 \, \text{N} \), acts upward, reducing the energy available for the firefighter to convert into kinetic energy. The work done by friction is calculated by the product of the friction force and the distance: \( f_s h \). This value is subtracted from the initial gravitational potential energy to get the amount of energy that converts to kinetic energy: \( mgh - f_s h = \frac{1}{2}mv^2 \). The reduced energy results in a slower speed as she reaches the ground, demonstrating the impact of the frictional force.

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Most popular questions from this chapter

The temperature of a plastic cube is monitored while the cube is pushed \(3.0 \mathrm{~m}\) across a floor at constant speed by a horizontal force of \(15 \mathrm{~N}\). The thermal energy of the cube increases by \(20 \mathrm{~J} .\) What is the increase in the thermal energy of the floor along which the cube slides?

A certain spring is found not to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .(\mathrm{b})\) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile-Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

A uniform cord of length \(25 \mathrm{~cm}\) and mass \(15 \mathrm{~g}\) is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\), where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\). with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and \((\mathrm{d})\) positive value of \(x\) is the potential energy equal to zero?
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