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Chapter 6: Problem 49

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

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

Step by step solution

01

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.
02

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 \(W_g = 80 \times 9.81 \times 3 \times \cos(0º) = 80 \times 9.81 \times 3 = 2354.4 \thinspace J\).
03

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 \(W_f = 400 \times 3 \times \cos(180º) = -1200 \thinspace J\).
04

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: \(W_{net} = W_g + W_f = 2354.4 - 1200 = 1154.4 \thinspace J\).
05

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.

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