Problem 61 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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 \(d\).

Step 3: 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.

Step 4: 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.

Step 5: Compute Specific Average Friction

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

Step 6: 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 \(d\).

Key Concepts

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.