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?

Short answer

Answer: The final speed of the fireman when he reaches the ground is 5.37 m/s.

Step-by-step solution

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.

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 \(W_g = 80 \times 9.81 \times 3 \times \cos(0º) = 80 \times 9.81 \times 3 = 2354.4 \thinspace J\).

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 \(W_f = 400 \times 3 \times \cos(180º) = -1200 \thinspace J\).

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: \(W_{net} = W_g + W_f = 2354.4 - 1200 = 1154.4 \thinspace J\).

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 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.

Understanding frictional force is crucial in determining the net work done on an object because it directly influences the object's motion. By applying a frictional force, the fireman is effectively reducing the speed at which he would slide under the influence of gravity alone.

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.

This force plays a significant role in determining the energy changes that occur as the fireman slides down the pole.

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.

By understanding the work-energy principle, one can calculate the speed or velocity of the fireman, knowing he starts from rest and only has the force of friction and gravity acting during his descent.

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.

The change in kinetic energy is what the work-energy principle uses to determine the fireman’s final speed. Since he starts from rest, his initial kinetic energy is zero, and therefore, his final kinetic energy is entirely due to the net work done during his slide.