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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) 9.19 m/s; (b) 4.94 m/s.

Step by step solution

01

Understand the Problem

For the firefighter sliding down without friction, we need to find the final velocity using the principles of energy conservation. In the second part, with friction involved, we need to calculate the final velocity using net force analysis.
02

Gather Known Values

Mass of firefighter ( m ) = 70 kg, height ( h ) = 4.3 m, gravitational acceleration ( g ) = 9.81 m/s虏. Frictional force when gripping = 500 N.
03

Calculate Velocity Without Friction

Use the conservation of energy principle. Initial Potential Energy = Final Kinetic Energy.\[ mgh = \frac{1}{2}mv^2 \]Cancel mass ( m ) and solve for velocity ( v ):\[ gh = \frac{1}{2}v^2 \]\[ v^2 = 2gh \]\[ v = \sqrt{2gh} = \sqrt{2 \cdot 9.81 \cdot 4.3} \]\[ v \approx 9.19 \text{ m/s} \]
04

Analyze Forces with Friction

When friction is involved, calculate net acceleration first. Use the formula:\[ F_{net} = ma = mg - F_{friction} \]\[ a = \frac{mg - F_{friction}}{m} \]\[ a = \frac{70 \times 9.81 - 500}{70} \]\[ a \approx 2.84 \text{ m/s}^{2} \]
05

Calculate Velocity with Friction

Use kinematic equations for constant acceleration:\[ v^2 = u^2 + 2a \cdot d \]Where initial velocity ( u ) = 0.\[ v^2 = 0 + 2 \times 2.84 \times 4.3 \]\[ v^2 = 24.424 \]\[ v \approx 4.94 \text{ m/s} \]

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental concept in physics. It asserts that energy cannot be created or destroyed, but only transformed from one form to another.
In the firefighter problem, energy conservation helps us understand how the potential energy at the top converts to kinetic energy at the bottom.
  • Potential Energy (PE): The firefighter starts at the top with a certain height, meaning she has gravitational potential energy calculated by the formula: \( PE = mgh \).
  • Kinetic Energy (KE): As she slides down, this potential energy is converted to kinetic energy, which can be expressed as \( KE = \frac{1}{2}mv^2 \).

The energy conservation principle allows us to set initial potential energy equal to the final kinetic energy for the frictionless case, \( mgh = \frac{1}{2}mv^2 \), ultimately allowing us to solve for her speed just before reaching the ground.
Kinematic Equations
Kinematic equations are invaluable tools in physics for solving problems involving motion, especially when considering constant acceleration.
The problem with the firefighter requires an understanding of these equations to determine speed both with and without friction.
  • No Friction: The final velocity of the firefighter can also be computed using the kinematic equation \( v^2 = u^2 + 2ad \), where \( u \) is the initial velocity (zero in this case), \( a \) is acceleration due to gravity, and \( d \) is the distance.
  • With Friction: If friction applies, the acceleration \( a \) changes and is determined by net forces.

These equations help trace the pathways of moving objects by predicting their future positions and velocities, allowing us to comprehensively solve motion problems.
Frictional Force
Frictional force acts opposite to the direction of motion, significantly altering the movement of objects. It's essential to consider friction when analyzing motion on surfaces or interactions, like the firefighter sliding down a pole. When the pole is held lightly, friction is negligible, simplifying calculations.
  • With negligible friction: The firefighter's path is governed solely by the force of gravity.
  • With friction: The force of 500 N works against the gravitational pull, thus reducing her speed. The presence of friction modifies the net force, changing both acceleration and velocity as a consequence.
  • Friction Formula: The net acceleration affected by friction can be calculated by subtracting the frictional force from the gravitational force \( F_{friction} \).

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Most popular questions from this chapter

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

The surface of the continental United States has an area of about \(8 \times 10^{6} \mathrm{~km}^{2}\) and an average elevation of about \(500 \mathrm{~m}\) (above sea level). The average yearly rainfall is \(75 \mathrm{~cm} .\) The fraction of this rainwater that returns to the atmosphere by evaporation is \(\frac{2}{3} ;\) the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water-Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg} .)\)

When a click beetle is upside down on its back, it jumps upward by suddenly arching its back, transferring energy stored in a muscle to mechanical energy. This launching mechanism produces an audible click, giving the beetle its name. Videotape of a certain clickbeetle jump shows that a beetle of mass \(m=4.0 \times 10^{-6} \mathrm{~kg}\) moved directly upward by \(0.77 \mathrm{~mm}\) during the launch and then to a maximum height of \(h=0.30 \mathrm{~m}\). During the launch, what are the average magnitudes of (a) the external force on the beetle's back from the floor and (b) the acceleration of the beetle in terms of \(g ?\)

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}(\) Fig. \(8-59) .\) A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m} ?\) (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m} ?\)

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm} .\) How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?\) (d) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?
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