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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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Most popular questions from this chapter

A runner reaches the top of a hill with a speed of \(6.50 \mathrm{~m} / \mathrm{s}\) He descends \(50.0 \mathrm{~m}\) and then ascends \(28.0 \mathrm{~m}\) to the top of the next hill. His speed is now \(4.50 \mathrm{~m} / \mathrm{s}\). The runner has a mass of \(83.0 \mathrm{~kg} .\) The total distance that the runner covers is \(400 . \mathrm{m}\) and there is a constant resistance to motion of \(9.00 \mathrm{~N}\). Use energy considerations to find the work done by the runner over the total distance.

A body of mass \(m\) moves in one dimension under the influence of a force, \(F(x)\), which depends only on the body's position. a) Prove that Newton's Second Law and the law of conservation of energy for this body are exactly equivalent. b) Explain, then, why the law of conservation of energy is considered to be of greater significance than Newton's Second Law.

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A 1.00 -kg block is pushed up and down a rough plank of length \(L=2.00 \mathrm{~m},\) inclined at \(30.0^{\circ}\) above the horizontal. From the bottom, it is pushed a distance \(L / 2\) up the plank, then pushed back down a distance \(L / 4,\) and finally pushed back up the plank until it reaches the top end. If the coefficient of kinetic friction between the block and plank is \(0.300,\) determine the work done by the block against friction.

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)
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