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 amounts of energy. initial: unit work? unit loss? unit final: unit what mass student would barely touch the water? unit would a student with a mass of 94kg hit the water? 145m 46m reference level check answers cannot be solved

Explanation:

Step1: Identify Energy Types and Changes

The problem involves gravitational potential energy (GPE) and elastic potential energy (EPE) for the bungee jump. Initially, the student has GPE (at height \( h = 145\,\text{m} \) above water, but wait, actually, the bridge height and cord length: wait, the natural length of the cord is \( 46\,\text{m} \)? Wait, no, let's re-express. Let's assume the initial state is at the bridge (height \( H \) above water), and the final state is when the student barely touches the water (height \( 0 \) above water). The cord stretches by \( x \), so the total fall distance is \( H = 46\,\text{m} + x \)? Wait, maybe the bridge height is \( 145\,\text{m} \), and the cord's natural length is \( 46\,\text{m} \), so the stretch \( x \) is what we need? Wait, no, the energy conservation: initial GPE (at bridge) = final EPE (when cord is stretched) + work? Wait, no, neglecting energy loss, initial energy (GPE) = final energy (EPE), because at the bottom (touching water), velocity is zero? Wait, no, when barely touching water, the kinetic energy is zero (momentarily at rest). So \( mgh = \frac{1}{2}kx^2 \), where \( h \) is the total fall distance (bridge height to water), and \( x \) is the stretch of the cord (beyond natural length). Wait, the bridge height is \( 145\,\text{m} \), natural length of cord is \( 46\,\text{m} \), so the stretch \( x = 145 - 46 = 99\,\text{m} \)? No, that can't be, because the spring constant is \( 24\,\text{N/m} \), which is small. Wait, maybe the initial mass is missing? Wait, the first question is "How much would the cord’s length change...", but maybe we need to find the mass first? Wait, no, the problem says "An engineer is trying to determine the mass of a student...". Wait, maybe the diagram shows bridge height \( 145\,\text{m} \), cord natural length \( 46\,\text{m} \), so the stretch \( x = 145 - 46 = 99\,\text{m} \)? No, that's too much. Wait, no, energy conservation: \( mgh = \frac{1}{2}kx^2 \), where \( h \) is the distance fallen (from bridge to water), and \( x \) is the stretch of the cord (so \( h = L + x \), where \( L = 46\,\text{m} \) is natural length). So \( mg(L + x) = \frac{1}{2}kx^2 \). But we need to find \( m \)? Wait, no, the first sub-question: "How much would the cord’s length change...", maybe we need to find \( x \) for a given mass? Wait, the problem is a bit unclear, but let's proceed with energy conservation.

Step2: Apply Energy Conservation (Simplified)

Assume initial energy: gravitational potential energy \( U_g = mgh \) (at bridge, height \( h = 145\,\text{m} \) above water). Final energy: elastic potential energy \( U_e = \frac{1}{2}kx^2 \) (when cord is stretched by \( x \), and kinetic energy is zero). So \( mgh = \frac{1}{2}kx^2 \). But we need to find \( m \) or \( x \)? Wait, the first question is "How much would the cord’s length change...", maybe for the student whose mass is such that they barely touch water. Wait, maybe the bridge height is \( 145\,\text{m} \), cord natural length \( 46\,\text{m} \), so the total fall distance is \( 145\,\text{m} \), so the stretch \( x = 145 - 46 = 99\,\text{m} \)? No, that's not right. Wait, maybe the initial height is \( 145\,\text{m} \) above the reference level (water), and the cord's natural length is \( 46\,\text{m} \), so when the student jumps, the cord stretches until the student reaches the water. So the distance fallen is \( 145\,\text{m} \), so the stretch \( x = 145 - 46 = 99\,\text{m} \). Then using \( mgh = \frac{1}{2}kx^2 \), solve for \( m \): \( m = \frac{kx^2}{2gh} \). Let's plug in \( k = 24\,\text{N/m} \), \( x = 99\,\text{m} \), \( g = 9.8\,\text{m/s}^2 \), \( h = 145\,\text{m} \). Wait, no, \( h \) is the height from bridge to water, which is the distance fallen, so \( h = 145\,\text{m} \), and the stretch \( x \) is the amount the cord stretches beyond its natural length, so \( x = h - L \), where \( L = 46\,\text{m} \), so \( x = 145 - 46 = 99\,\text{m} \). Then \( m = \frac{kx^2}{2gh} = \frac{24 \times 99^2}{2 \times 9.8 \times 145} \). Let's calculate that: \( 99^2 = 9801 \), \( 24 \times 9801 = 235224 \), \( 2 \times 9.8 \times 145 = 2842 \), so \( m = \frac{235224}{2842} \approx 82.8\,\text{kg} \). Wait, but the next question is "What mass student would barely touch the water?" So maybe that's the mass. Then "Would a student with mass 94kg hit the water?" Since 94kg > 82.8kg, the 94kg student would have more GPE, so the cord would stretch more, meaning they would hit the water (since the stretch would exceed the bridge height - cord length? Wait, no, if mass is larger, \( mgh = \frac{1}{2}kx^2 \) implies \( x \) is larger, so if \( x \) exceeds the distance from bridge to water (145m - 46m = 99m), then they hit the water. Wait, no, if \( m = 94\,\text{kg} \), then \( x = \sqrt{\frac{2mgh}{k}} = \sqrt{\frac{2 \times 94 \times 9.8 \times 145}{24}} \). Calculate numerator: 2×94×9.8×145 = 2×94×1421 = 188×1421 = 267148. Then \( x = \sqrt{\frac{267148}{24}} = \sqrt{11131.17} \approx 105.5\,\text{m} \). Then total fall distance is \( L + x = 46 + 105.5 = 151.5\,\text{m} \), which is more than the bridge height (145m), so the student would hit the water.

Step1 (Revised): Energy Conservation Setup

For the student barely touching water, initial energy (GPE at bridge) = final energy (EPE at maximum stretch, KE = 0). Let \( m \) be the student's mass, \( H = 145\,\text{m} \) (bridge height to water), \( L = 46\,\text{m} \) (cord natural length), \( k = 24\,\text{N/m} \), \( g = 9.8\,\text{m/s}^2 \). The stretch \( x = H - L = 145 - 46 = 99\,\text{m} \)? No, wait, \( H \) is the total fall distance, so \( x = H - L \) only if the cord stretches to the water. But actually, \( H = L + x \), so \( x = H - L \). Then \( mgh = \frac{1}{2}kx^2 \) (where \( h = H \), since GPE is \( mgh \) with \( h = H \)).

Step2: Calculate Mass for Barely Touching Water

Using \( m = \frac{kx^2}{2gH} \), \( x = 99\,\text{m} \), \( H = 145\,\text{m} \), \( k = 24\,\text{N/m} \), \( g = 9.8\,\text{m/s}^2 \):
\[ m = \frac{24 \times 99^2}{2 \times 9.8 \times 145} = \frac{24 \times 9801}{2842} \approx \frac{235224}{2842} \approx 82.8\,\text{kg} \]

Step3: Analyze 94kg Student

For \( m = 94\,\text{kg} \), calculate the required stretch \( x' \) from \( m'gh = \frac{1}{2}kx'^2 \):
\[ x' = \sqrt{\frac{2m'gh}{k}} = \sqrt{\frac{2 \times 94 \times 9.8 \times 145}{24}} = \sqrt{\frac{267148}{24}} \approx \sqrt{11131.17} \approx 105.5\,\text{m} \]
Total fall distance for 94kg: \( L + x' = 46 + 105.5 = 151.5\,\text{m} \), which is greater than the bridge height (145m), so the 94kg student would hit the water.

Answer:

• Mass of student barely touching water: \(\approx 82.8\,\text{kg}\)
• 94kg student: Yes, they would hit the water.

(Note: The exact calculations depend on precise values, but the key is energy conservation between GPE and EPE, and comparing the required stretch for 94kg to the available bridge height - cord length.)