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A 68.0 -kg bungee jumper is standing on a tall platform \(\left(h_{0}=\right.\) \(46.0 \mathrm{m}),\) as indicated in the figure. The bungee cord has a natural length of \(L_{0}=9.00 \mathrm{m}\) and, when stretched, behaves like an ideal spring with a spring constant of \(k=66.0 \mathrm{N} / \mathrm{m}\). The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Concepts: (i) Can we use the conservation of mechanical energy to find his speed at any point along the descent? Explain your answer. (ii) What type of energy does he have when he is standing on the platform? (iii) What types of energy does he have at point A? (iv) What types of energy does he have at point \(\mathrm{B} ?\) Calculations: What is his speed when he is at the following heights above the water: (a) \(h_{\mathrm{A}}=37.0 \mathrm{m},\) and (b) \(h_{\mathrm{B}}=15.0 \mathrm{m} ?\)

Step by step solution

01

Initial Energy Considerations

The jumper initially only has gravitational potential energy since he is standing on the platform. His initial potential energy can be calculated using the formula: \[ PE_{initial} = mgh_0 \] where \( m = 68.0 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( h_0 = 46.0 \text{ m} \).

02

Energy at Height h_A

At height \(h_A = 37.0 \text{ m}\), the jumper has both gravitational potential energy \( PE_A \) and kinetic energy \( KE_A \). At this point:\[ PE_A = mgh_A = 68 \times 9.81 \times 37 = 24664.44 \text{ J} \]Since the bungee cord is not yet stretched, there's no elastic potential energy. The total mechanical energy at point \( A \) is: \[ E_{total} = PE_{initial} = KE_A + PE_A \].Solving for the kinetic energy:\[ KE_A = PE_{initial} - PE_A \].

03

Calculate Speed at h_A

Use the kinetic energy found to determine the speed at height \(h_A\):\[ KE_A = \frac{1}{2}mv_A^2 \]\[ v_A = \sqrt{\frac{2KE_A}{m}} \].Substitute the values:\[ KE_A = 30723.6 - 24664.44 = 6069.16 \text{ J} \]\[ v_A = \sqrt{\frac{2 \times 6069.16}{68}} = 13.39 \text{ m/s} \].

04

Energy at Height h_B

At height \( h_B = 15.0 \text{ m}\), the jumper has gravitational potential energy, kinetic energy, and elastic potential energy since the cord is stretched. First, find the stretched length of the bungee cord:\[ L_{stretched} = h_0 - h_B = 46 - 15 = 31 \text{ m} \]. The stretched amount is:\[ x = L_{stretched} - L_0 = 31 - 9 = 22 \text{ m}\]. The elastic potential energy \(U_s\) is:\[ U_s = \frac{1}{2}kx^2 = \frac{1}{2} \times 66 \times 22^2 \].

05

Calculate Speed at h_B

The total mechanical energy at point B can be equated from the initial energy:\[ PE_{initial} = PE_B + KE_B + U_s \].Gravitational potential energy at \(h_B\):\[ PE_B = mgh_B = 68 \times 9.81 \times 15 = 10017 \text{ J} \].Elastic potential energy:\[ U_s = 15972 \text{ J} \].Substitute to find \(KE_B\):\[ KE_B = PE_{initial} - PE_B - U_s\].Kinetic energy formula:\[ v_B = \sqrt{\frac{2KE_B}{m}} \].Solve for the values:\[ KE_B = 30723.6 - 10017 - 15972 = 4744.6 \text{ J} \]\[ v_B = \sqrt{\frac{2 \times 4744.6}{68}} = 11.8 \text{ m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy

When the bungee jumper is standing on the platform, he possesses gravitational potential energy because of his height above the ground. This energy is potential because it has the potential to be converted into other forms of energy, like kinetic energy, as the jumper falls.

Gravitational potential energy is calculated using the formula:

• \[ PE = mgh \]

where:

• \( m \) is the mass of the jumper,
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and
• \( h \) is the vertical height of the jumper above a certain reference point.

Gravitational potential energy is highest when the jumper is on the platform before jumping. At this height (\( h_0 = 46.0 \, \text{m} \)), the gravitational potential energy is maximized, given by:

• \[ PE_{initial} = mgh_0 \]

This energy will decrease as gravitational potential is converted to kinetic energy during the descent.

Kinetic Energy

Kinetic energy is the energy of motion. As the jumper begins to fall from the platform, his gravitational potential energy starts converting into kinetic energy. This conversion happens because as the jumper falls, he gains speed.

The kinetic energy can be calculated using the formula:

• \[ KE = \frac{1}{2}mv^2 \]

where:

• \( m \) is the mass of the jumper,
• \( v \) is the velocity of the jumper.

As the jumper reaches **point A (37.0 m)**, he has a combination of gravitational potential energy and kinetic energy, but not yet elastic potential energy as the bungee cord hasn’t stretched beyond its natural length. The velocity at any given point can be calculated by deriving kinetic energy from the conservation of total mechanical energy.

• At point A:\[ KE_A = PE_{initial} - PE_A \]

Using this kinetic energy, the speed at height \( h_A \) can be found:

• \[ v_A = \sqrt{\frac{2KE_A}{m}} \]

Thus, kinetic energy plays a crucial role in determining how fast the jumper is falling at any point.

Elastic Potential Energy

Elastic potential energy comes into play when the bungee cord starts stretching. This happens as the jumper descends below the natural length of the bungee cord. The cord behaves like a spring and stores energy as it stretches.

The elastic potential energy can be calculated with:

• \[ U_s = \frac{1}{2}kx^2 \]

where:

• \( k \) is the spring constant of the bungee cord, and
• \( x \) is the extension of the bungee cord from its natural length.

At **point B (15.0 m)**, the jumper has all three types of energy: gravitational potential, kinetic, and elastic potential energy. The cord is fully stretched here, hence a significant portion of the gravitational potential energy has been converted into elastic potential energy and some into kinetic energy. Calculating the elastic potential energy at this point involves:

• \[ x = L_{stretched} - L_0 \]

and then using:

• \[ U_s = \frac{1}{2}kx^2 \]

This elastic energy is critical as it decides how much further the jumper will keep descending before bouncing back up.