Problem 49
Question
A 80.0 -kg fireman slides down a 3.00 -m pole by applying a frictional force of \(400 .\) N against the pole with his hands. If he slides from rest, how fast is he moving once he reaches the ground?
Step-by-Step Solution
Verified Answer
Answer: The final speed of the fireman when he reaches the ground is 5.37 m/s.
1Step 1: Identify the forces acting on the fireman and their respective work
The two forces acting on the fireman are gravity and friction. The gravitational force acts downward and has a magnitude equal to the weight of the fireman, which is W = mg, where m is the mass of the fireman (80 kg) and g is the acceleration due to gravity (9.81 m/s²). The frictional force acts upward and has a magnitude of 400 N (given in the problem statement). To calculate the work done by each force, we use the work formula: work = force x distance x cos(theta), where theta is the angle between the force and the displacement.
2Step 2: Calculate the work done by gravitational force
The gravitational force acts downward and the displacement (3 m) is also downward, so the angle between these two quantities is 0º. Therefore, the work done by gravity on the fireman is: \(W_g = mgd \times \cos(0º)\), where m = 80 kg, g = 9.81 m/s², and d = 3 m. Plugging in the values, we get \(W_g = 80 \times 9.81 \times 3 \times \cos(0º) = 80 \times 9.81 \times 3 = 2354.4 \thinspace J\).
3Step 3: Calculate the work done by frictional force
The frictional force acts upward and the displacement is downward, so the angle between these two quantities is 180º. Therefore, the work done by the frictional force on the fireman is: \(W_f = Fd \times \cos(180º)\), where F = 400 N and d = 3 m. Plugging in the values, we get \(W_f = 400 \times 3 \times \cos(180º) = -1200 \thinspace J\).
4Step 4: Calculate the net work done on the fireman
To find the net work done on the fireman, we add the work done by each force: \(W_{net} = W_g + W_f = 2354.4 - 1200 = 1154.4 \thinspace J\).
5Step 5: Use the work-energy principle to find the final speed of the fireman
According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. Since the fireman starts from rest, the initial kinetic energy is 0, and the final kinetic energy is \(\frac{1}{2}mv_f^2\), where \(v_f\) is the final speed. Therefore, we can write the equation: \(W_{net} = \frac{1}{2}mv_f^2\). Plugging in the values, we get \(1154.4 = \frac{1}{2}(80)v_f^2\). Solving for \(v_f^2\), we get \(v_f^2 = \frac{1154.4 \times 2}{80} = 28.86\). Taking the square root of both sides, we find the final speed: \(v_f = \sqrt{28.86} = 5.37 \thinspace m/s\).
So, the fireman is moving at a speed of 5.37 m/s once he reaches the ground.
Key Concepts
Frictional ForceGravitational ForceWork-Energy PrincipleKinetic Energy
Frictional Force
Frictional force is an important concept in physics that comes into play whenever two surfaces are in contact with each other. In our problem, the fireman uses his hands to apply a frictional force against the pole as he slides down. This force opposes the motion of the fireman, acting upward while he moves downward.
- Frictional forces are parallel to the contact surfaces.
- It acts to resist the relative motion or tendency of such motion of the surfaces.
- In the problem, it has a magnitude of 400 N, which means it applies an upward force of that magnitude.
Gravitational Force
Gravitational force is the force of attraction between two masses. Here, it acts as the force pulling the fireman downward towards the center of the Earth. The magnitude of this force is equal to the weight of the fireman, calculated using the formula: \( W = mg \), where \( m \) is the mass of the fireman and \( g \) is the acceleration due to gravity.
- For the fireman, \( m = 80 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \).
- Gravitational force acts vertically downwards, influencing the fireman's downward slide.
- The work done by this force is given by \( W_g = mgd \times \cos(0º) \), as the force and displacement are in the same direction.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the work done by forces on an object to a change in the object's kinetic energy. In simple terms, it states that the work done on the system is equal to the change in kinetic energy of that system.
- This principle is crucial to solving problems where different forces act on moving objects.
- It's defined mathematically as \( W_{\text{net}} = \Delta K = K_f - K_i \), where \( K_f \) and \( K_i \) are the final and initial kinetic energies, respectively.
- In the given scenario, this principle helps determine how fast the fireman is moving when he hits the ground by using the work done by both gravitational and frictional forces.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on two factors: the mass of the object and its velocity. The formula for kinetic energy is \( K = \frac{1}{2}mv^2 \).
- For objects starting from rest, the initial kinetic energy is zero because speed is zero.
- As the fireman slides down, his speed increases, and so does his kinetic energy.
- At the bottom of the pole, the kinetic energy of the fireman has increased due to the net work done by both gravitational and frictional forces.
Other exercises in this chapter
Problem 46
A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1
View solution Problem 47
A horizontal slingshot consists of two light, identical springs (with spring constants of \(30.0 \mathrm{~N} / \mathrm{m}\) ) and a light cup that holds a 1.00
View solution Problem 50
A large air-filled 0.100 -kg plastic ball is thrown up into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\). At a height of \(3.00 \mathrm{~m
View solution Problem 51
How much mechanical energy is lost to friction if a 55.0-kg skier slides down a ski slope at constant speed of \(14.4 \mathrm{~m} / \mathrm{s}\) ? The slope is
View solution