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Chapter 7: Problem 61

Down the Pole. A fireman of mass \(m\) slides a distance \(d\) down a pole. He starts from rest. He moves as fast at the bottom as if be 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 \(75-\mathrm{kg}\) fireman exerts, for \(d=2.5 \mathrm{m}\) and \(h=1.0 \mathrm{m},(\mathrm{c})\) 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) The average friction force is \( f_{\text{friction}} = \frac{mg(d-h)}{d} \), (b) 441.45 N, (c) \(v = \sqrt{2g(h-y)}\).

Step by step solution

01

Understand the Scenario

The fireman slides down a pole from rest, starting at height \(d\), with apparent height \(h\) for free fall with negligible air resistance. Average friction force is the concern needed to bring his speed down to actual speed to match free fall.
02

Apply Conservation of Energy

When sliding, the difference between positions at \(h\) and \(d\) measures the energy transferred due to sliding. This is expressed as: \( mg(d - h) = f_{\text{friction}} \cdot d \). The term \(mgd\) converts potential energy to an equivalent force working over distance slide \(d\).
03

Calculate Average Friction Force

Solve for the average friction: \[ f_{\text{friction}} = \frac{mg(d-h)}{d} \]. This calculation relates differences in assumed potential energy fall height \(h\) relative to height \(d\) considering friction work.
04

Check Consistency with Special Cases

If \(h=d\), \(d-h=0\) means the fireman slides freely without friction which is consistent intuitively, solving results: \( f_{\text{friction}}=0\). While for \(h=0\), \( f_{\text{friction}} = mg\) indicating full energy dissipated through friction.
05

Compute Specific Average Friction

Plugging in given values \(m=75 \text{ kg}, d=2.5 \text{ m}, h=1.0 \text{ m}\): \[ f_{\text{friction}} = \frac{75 \times 9.81 (2.5 - 1)}{2.5} = 441.45 \text{ N} \].
06

Determine Speed at Distance y Above Bottom

The speed, with starting at rest, is derived from \(c=0\) using energy ideas: \[ v^2 = 2g(h - y ) \]. When potential difference \(hc=h-y\) progresses with time, instead of simple position \(d\).

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Key Concepts

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

Friction Force
When a fireman slides down a pole, there is contact between him and the pole, creating a force known as friction. Friction acts against the direction of motion, working to slow down the fireman's descent. This force is crucial in ensuring safety, as it prevents the fireman from descending too quickly and potentially injuring himself at the bottom.
Friction force in this scenario can be expressed in terms of energy transfer. As the fireman descends, he loses potential energy, which is transformed into kinetic energy and some of that energy is dissipated through friction. This friction force can be calculated using the equation: \[ f_{\text{friction}} = \frac{mg(d-h)}{d} \]where:
  • \( m \) is the mass of the fireman
  • \( g \) is the acceleration due to gravity
  • \( d \) is the distance slid down the pole
  • \( h \) is the height simulating free fall
This relationship highlights how the difference in potential energy from heights \( h \) and \( d \) affects the average friction force acting on the fireman.
Free Fall
Free fall is a fascinating concept in physics, referring to an object's motion when gravity is the only force acting upon it. In the scenario involving the fireman, if he jumped without friction, he would experience free fall. In this case, his descent velocity would increase solely due to gravity, starting from rest.
However, in a realistic situation, free fall isn't entirely applicable as friction with the pole plays a significant role. But the problem uses the free fall concept to equate the fireman's descent with a hypothetical drop from height \( h \).
The equivalent speed of the fireman as if he were in free fall is determined by assuming that all potential energy is converted to kinetic energy without friction. This perfect scenario is given by the potential height \( h \), which is a measure of how far he could "fall" if there were no obstructions such as friction.
Potential Energy
Potential energy is the energy stored in an object due to its position or state. For the fireman sliding down the pole, his potential energy at the topmost point is linked to his height above the ground. This energy can be defined as gravitational potential energy, which depends on several factors:
The equation for gravitational potential energy is:\[ PE = mgh \]where:
  • \( PE \) is the potential energy
  • \( m \) is the mass of the fireman
  • \( g \) is the acceleration due to gravity
  • \( h \) is the height from which he starts
As the fireman descends, his potential energy decreases, being converted into kinetic energy. It is also partly balanced by the friction force acting against his motion. Understanding how potential energy transforms into other forms of energy, like kinetic energy and dissipative forces such as friction, is crucial for solving problems involving energy conservation.
Kinetic Energy
Kinetic energy is the energy of motion. As the fireman slides down the pole, his potential energy is transferred to kinetic energy, causing him to accelerate downward. The amount of kinetic energy can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]In this case:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the fireman
  • \( v \) is the velocity of the fireman as he slides
The fireman's kinetic energy increases as he loses height and gains speed. The difference between the initial potential energy at height \( h \) and the kinetic energy acquired during the descent equates to the work done by friction. As the fireman nears the ground, his kinetic energy becomes maximal just before being dissipated at the bottom of the pole.

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Most popular questions from this chapter

Gravity in Three Dimensions. A point mass \(m_{1}\) is held in place at the origin, and another point mass \(m_{2}\) is free to move a distance \(r\) away at a point \(P\) having coordinates \(x, y,\) and \(z\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant (see Exercises 7.34 and 7.35 . (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ $$ F_{z}=-\frac{G m_{1} m_{2} z}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x, y,\) and \(z\) ) (b) Show that the magnitude of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\left(\text { c) Does } m_{1} \text { attract or }\right.\) repel \(m_{2} ?\) How do you know?

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 particle with mass \(m\) is acted on by a conservative force and moves along a path given by \(x=x_{0} \cos \omega_{0} t\) and \(y=y_{0} \sin \omega_{0} t\) where \(x_{0}, y_{0},\) and \(\omega_{0}\) are constants. (a) Find the components of the force that acts on the particle. (b) Find the potential energy of the particle as a function of \(x\) and \(y .\) Take \(U=0\) when \(x=0\) and \(y=0 .\left(\text { c) Find the total energy of the particle when (i) } x=x_{0 .}\right.\) \(y=0\) and \(\left(\text { ii) } x=0, y=y_{0}\right.\)

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{\boldsymbol{F}}=-\alpha x^{2} \hat{i},\) where \(\alpha=12 \mathrm{N} / \mathrm{m}^{2} .\) (a) How much work does \(\overrightarrow{\boldsymbol{F}}\) do when the proton moves along the straight- line path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0.40 \mathrm{m}) ?(\mathrm{b})\) Along the straightline path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.30 \mathrm{m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0) ?(\mathrm{d})\) Is the force \(\overrightarrow{\boldsymbol{F}}\) conservative? Explain. If \(\overrightarrow{\boldsymbol{F}}\) is conservative, what is the potential energy function for it? Let \(U=0\) when \(x=0\) .

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.
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