Question

an engineer is trying to determine the mass of a student who will barely touch the water when jumping off a bridge shown with an ideal bungee cord with a spring constant of 24n/m. the height of the bridge and natural length of the bungee cord are shown. neglect drag, energy gained from the jump and loss due to impact with the water.
how much would the cord’s length change in order to touch the water?
unit
list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amount of energy.
initial:
=
unit
work?
=
unit
loss?
=
unit
final:
=
unit
(there is a diagram at the bottom showing a bridge height of 145m, the natural length of the bungee cord as 46m, and the student’s position at jumping and at the lowest point near the reference level (water).)

Explanation:

Step1: Identify Known Values

The bridge height is \( 145 \, \text{m} \), natural length of bungee cord \( L_0 = 46 \, \text{m} \), spring constant \( k = 24 \, \text{N/m} \). Let the mass be \( m \), and at the lowest point (touching water), the student is at rest (final velocity \( v_f = 0 \)), initial velocity \( v_i = 0 \).

Step2: Energy Conservation (Initial to Final)

Initial energy: Gravitational Potential Energy (GPE) \( U_i = mgh_i \), Kinetic Energy (KE) \( K_i = \frac{1}{2}mv_i^2 = 0 \), Elastic Potential Energy (EPE) \( E_i = 0 \) (cord not stretched initially).
Final energy: GPE \( U_f = mgh_f \), KE \( K_f = 0 \), EPE \( E_f = \frac{1}{2}k\Delta x^2 \), where \( \Delta x = h - L_0 \) (stretch of cord, \( h \) is total fall distance = bridge height? Wait, no—wait, the student jumps, so the total fall distance is the distance from bridge to water, let's say the lowest point is at water level, so the displacement from bridge to water is \( 145 \, \text{m} \). The cord stretches by \( \Delta x = 145 - 46 = 99 \, \text{m} \)? Wait, no, initial position: student is at bridge, cord length \( 46 \, \text{m} \) (natural length). When jumping, the cord stretches until the student touches water. So the total fall distance is the length the cord stretches plus its natural length? Wait, no—bridge height is \( 145 \, \text{m} \), the student’s initial position (at bridge) has height \( h_i = 145 \, \text{m} \) (reference level at water). Final position: \( h_f = 0 \). The cord’s stretch \( \Delta x = \) total fall distance \( - \) natural length \( = 145 - 46 = 99 \, \text{m} \).

Initial energy: \( E_{\text{total},i} = U_i + K_i + E_i = m g (145) + 0 + 0 = 145mg \).
Final energy: \( E_{\text{total},f} = U_f + K_f + E_f = 0 + 0 + \frac{1}{2}k(99)^2 \).

By energy conservation (neglecting drag, so no non-conservative work), \( E_{\text{total},i} = E_{\text{total},f} \)? Wait, no—wait, the problem says "energy gained from the jump and loss due to impact"—maybe I misread. Wait, the engineer wants to find the mass. Wait, the student "barely touches the water", so at the lowest point, velocity is zero. So initial KE = 0, final KE = 0. Initial EPE = 0 (cord not stretched), final EPE = \( \frac{1}{2}k\Delta x^2 \), where \( \Delta x = h - L_0 \), \( h \) is the distance fallen (bridge height? Wait, the diagram shows bridge height 145m, and a 46m segment. So the natural length is 46m, so when the student jumps, the cord stretches by \( x = 145 - 46 = 99 \, \text{m} \) (since the bridge height is 145m, and the cord’s natural length is 46m, so the stretch is total fall minus natural length).

Using conservation of mechanical energy (KE + GPE + EPE = constant, since no non-conservative forces, as drag is neglected):
\( K_i + U_i + E_i = K_f + U_f + E_f \)
\( 0 + m g (145) + 0 = 0 + 0 + \frac{1}{2}k(99)^2 \)

Wait, but we need to find mass? Wait, no—the question is "How much would the cord’s length change in order to touch the water?" Wait, maybe the bridge height is 145m, and the natural length is 46m, so the stretch is \( 145 - 46 = 99 \, \text{m} \), so the cord’s length change (stretch) is 99m. Wait, but maybe I messed up. Wait, the initial position: student at bridge, cord length 46m (natural). Final position: student at water, cord length \( 46 + \Delta x \), and the total distance from bridge to water is 145m, so \( 46 + \Delta x = 145 \), so \( \Delta x = 145 - 46 = 99 \, \text{m} \). So the cord’s length changes by 99 meters (stretches by 99m).

For the energy types:
- Initial:
- KE: 0 (starts from rest), unit: J
- GPE: \( mgh \) (h = 145m), unit: J
- EPE: 0 (cord not stretched), unit: J
- Work: 0 (no non-conservative work initially), unit: J
- Loss: 0 (no loss initially), unit: J

- Final:
- KE: 0 (stops at water), unit: J
- GPE: 0 (at reference level), unit: J
- EPE: \( \frac{1}{2}k\Delta x^2 = \frac{1}{2} \times 24 \times 99^2 \), unit: J
- Work: 0 (no non-conservative work during ideal jump), unit: J
- Loss: 0 (neglect drag, so no loss), unit: J

Wait, but the problem says "energy gained from the jump and loss due to impact"—maybe the initial energy is the energy from the jump (KE initial? No, initial KE is 0). Wait, maybe the student jumps with initial velocity, but the problem says "ideal bungee cord" and "neglect drag, energy gained from the jump"—maybe "energy gained" is the GPE converted to EPE. Wait, perhaps the mass is found first. Let's solve for mass:

From energy conservation:
\( m g (145) = \frac{1}{2} \times 24 \times (99)^2 \)
\( m = \frac{24 \times 99^2}{2 \times 9.8 \times 145} \)
Calculate \( 99^2 = 9801 \)
\( 24 \times 9801 = 235224 \)
\( 2 \times 9.8 \times 145 = 2 \times 1421 = 2842 \)
\( m = \frac{235224}{2842} \approx 82.76 \, \text{kg} \) (approx 83kg).

But the main question here is "How much would the cord’s length change in order to touch the water?" The cord’s natural length is 46m, and the total fall distance is 145m, so the stretch (length change) is \( 145 - 46 = 99 \, \text{m} \).

For the energy table:
- Initial:
- KE: 0 J (starts at rest)
- GPE: \( m \times 9.8 \times 145 \) J (depends on mass, but if we take m ≈ 83kg, \( 83 \times 9.8 \times 145 ≈ 83 \times 1421 ≈ 117943 \) J)
- EPE: 0 J (cord not stretched)
- Work: 0 J (no non-conservative forces yet)
- Loss: 0 J (no loss yet)

- Final:
- KE: 0 J (stops at water)
- GPE: 0 J (at water level)
- EPE: \( \frac{1}{2} \times 24 \times 99^2 = 12 \times 9801 = 117612 \) J
- Work: 0 J (ideal, no non-conservative work)
- Loss: 0 J (neglect drag)

Wait, the GPE initial and EPE final are almost equal (difference due to rounding mass), which makes sense for energy conservation (KE initial and final are zero).

Answer:

The cord’s length changes (stretches) by \( \boldsymbol{99 \, \text{meters}} \) to touch the water.

For the energy table (example with \( m \approx 83 \, \text{kg} \)):

• Initial:
• KE: \( 0 \, \text{J} \)
• GPE: \( \approx 117943 \, \text{J} \)
• EPE: \( 0 \, \text{J} \)
• Work: \( 0 \, \text{J} \)
• Loss: \( 0 \, \text{J} \)
• Final:
• KE: \( 0 \, \text{J} \)
• GPE: \( 0 \, \text{J} \)
• EPE: \( 117612 \, \text{J} \)
• Work: \( 0 \, \text{J} \)
• Loss: \( 0 \, \text{J} \)

(Note: The slight difference in GPE and EPE is due to rounding the mass; with exact calculation, \( m = \frac{24 \times 99^2}{2 \times 9.8 \times 145} = \frac{235224}{2842} ≈ 82.76 \, \text{kg} \), so GPE \( = 82.76 \times 9.8 \times 145 ≈ 117612 \, \text{J} \), matching EPE, confirming energy conservation.)