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Chapter 7: Problem 61

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He sturts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance \(h \leq d\) above the ground and descended with negligible air resistance. (a) What average friction force did the fireman exert on the pole? Does your answer make sense in the special cases of \(h=d\) and \(h=0 ?\) (b) Find a numerical value for the average friction force a 75 -kg fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m}\) (c) In terms of \(g, h,\) and \(d,\) what is the speed of the fireman when he is a distance \(y\) above the bottom of the pole?

Short Answer

Expert verified
(a) \(f_f = mg(d - h)\). (b) 1102.5 N. (c) \(v = \sqrt{2g(h - y)}\).

Step by step solution

01

Understanding the Problem

We need to find the average friction force exerted by a fireman sliding down a pole. The fireman's speed at the bottom is the same as if he free-fell from height \(h\) without air resistance. We'll first use energy conservation concepts to find the friction force and then check if this makes sense for given special cases.
02

Applying Energy Conservation

When the fireman slides down, his potential energy changes to kinetic energy and work done against friction. We have:\[ mgh_f = mgh + f_f \cdot d \]where \( h_f \) is the final height, \( h \) is the equivalent free-fall height, and \( f_f \) is the frictional force. Set \(h_f = 0\) to solve for \(f_f\).
03

Solving for Friction Force

Rearrange the conservation equation to isolate the frictional force \(f_f\):\[ f_f = mg(d - h) \]
04

Special Case: \(h = d\)

If \( h = d \), the fireman experiences free fall for the entire distance without friction, hence \(f_f = 0\). This makes sense because no energy is lost to friction.
05

Special Case: \(h = 0\)

If \( h = 0 \), the fireman experiences maximum friction, converting all initial potential energy to work against friction:\[ f_f = mgd \]This also makes sense as the energy is entirely used to overcome friction.
06

Plugging in Numbers

Using \(m = 75 \text{ kg}, d = 2.5 \text{ m}, h = 1.0 \text{ m}\), calculate:\[ f_f = 75 \cdot 9.8 \cdot (2.5 - 1.0) = 75 \cdot 9.8 \cdot 1.5 = 1102.5 \text{ N} \]
07

Speed at Distance \(y\) Above Ground

The energy conservation equation at height \(y\) is:\[ mgh = \frac{1}{2}mv^2 + f_f \cdot (d - y) \]Solve for speed \(v\) using:\[ v = \sqrt{2g(h - y)} \]

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Key Concepts

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

Friction Force
When the fireman slides down the pole, he exerts an average friction force on the pole. This force is crucial because it opposes the motion and transforms some of the potential energy that would otherwise translate into kinetic energy. The friction force can be thought of as the 'braking' factor in the scenario, preventing the fireman from accelerating as freely as he would in a non-resistant fall. In the context of our problem, the average friction force can be calculated using the derived formula: - Rearrange energy conservation to find the force: - Substitute known values to compute the exact frictional force.
Kinetic Energy
Kinetic energy is the energy of motion. For the fireman, as he accelerates down the pole, he converts potential energy (from his height above ground) into kinetic energy. - It is given by: - Initially starting from rest means he has zero kinetic energy. - As he descends, his speed increases, reducing potential energy, while his kinetic energy simultaneously increases. - The overall energy within the system remains constant, only converting between forms, provided no other external work is introduced, other than friction.
Potential Energy
Potential energy is the stored energy due to an object's position. In this exercise, it refers to the fireman's initial height above the ground.- His potential energy is calculated using: - At the top, this energy is maximum and gradually decreases as he descends down the pole.- As he comes closer to the ground, his height decreases, effectively reducing his potential energy and turning it into kinetic energy and work against friction.- The significance of determining equivalent free-fall height \( h \) is essential because it sets an energy baseline, acknowledging the work done by friction for an equivalent descent.
Equations of Motion
Equations of motion connect the dots between forces, motions, and energy transformations. They allow us to predict subsequent positions and velocities of the fireman as he slides. - By applying the principles of energy conservation, and including friction, these equations tell us how fast the fireman will be moving at any point.- The energy conservation equation becomes: - Solving this helps find the velocity of the fireman as distance \( y \) from the ground is changed: - With this equation, students can calculate speed alterations as the fireman descends and assess how height and friction impact these velocities.- This provides a real-world appreciation for the interplay between forces in motion and is essential to understanding energy conservation in dynamic systems.

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Most popular questions from this chapter

CALC An object has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=\alpha x y \hat{\boldsymbol{r}},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (Sce Problem \(6.98 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2}\) . The object moves along the following path: \((1)\) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\vec{F}\) for each leg of the path and for the complete round trip. (c) Is \(\vec{F}\) conservative or nonconservative? Explain.

EALE A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m}\) ?

A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m} .\) The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

A 75 -kg rooter climbs a vertical 7.0 -m ladder to the flat roof of a house. He then walks 12 \(\mathrm{m}\) on the roof, climbs down another vertical 7.0 -m ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; (b) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? (e) On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.
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