Skip to main content
Ch. 08 - Conservation of Energy
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 8, Problem 24

Chris jumps off a bridge with a 15-m-long bungee cord (a heavy stretchable cord) tied around his ankle, Fig. 8–37. He falls 15 m before the bungee cord begins to stretch. Chris’s mass is 75 kg and we assume the cord obeys Hooke’s law, F = -kx with k = 55 N/m. If we neglect air resistance, estimate what distance d below the bridge Chris’s foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle.
Illustration showing a person jumping off a bridge with a bungee cord, depicting fall distance and elastic potential energy.

Verified step by step guidance
1
Step 1: Break the motion into two parts: (a) the free fall of 15 m before the bungee cord starts to stretch, and (b) the motion where the bungee cord stretches and applies a restoring force according to Hooke's law. Analyze each part separately.
Step 2: Calculate the velocity of Chris just before the bungee cord starts to stretch using the kinematic equation for free fall: v=u+gt. Here, u is the initial velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the time of free fall. Alternatively, use the energy conservation principle to find the velocity.
Step 3: Once the bungee cord starts to stretch, the system involves both gravitational potential energy and elastic potential energy. Write the total energy conservation equation: E=mgh+12kx=2, where h is the total distance fallen, k is the spring constant, and x is the extension of the cord.
Step 4: Solve for the total distance d below the bridge by combining the 15 m free fall and the additional stretch of the cord. Use the quadratic equation derived from the energy conservation equation to find x, the stretch of the cord.
Step 5: Add the 15 m free fall distance to the calculated stretch of the cord to determine the total distance d below the bridge where Chris comes to a stop. Ensure all units are consistent throughout the calculations.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a spring (or elastic material) is directly proportional to the amount it is stretched or compressed, represented mathematically as F = -kx. Here, F is the restoring force, k is the spring constant, and x is the displacement from the equilibrium position. This principle is crucial for understanding how the bungee cord behaves as Chris falls and begins to stretch.
Recommended video:
Guided course
05:27
Spring Force (Hooke's Law)

Gravitational Force

The gravitational force acting on an object is the force due to gravity, calculated using F = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth). In this scenario, Chris's weight contributes to the downward force during his fall, which must be balanced by the upward force from the bungee cord when it stretches.
Recommended video:
Guided course
05:41
Gravitational Forces in 2D

Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In this context, as Chris falls, gravitational potential energy is converted into elastic potential energy when the bungee cord stretches. Understanding this energy transformation is essential for calculating how far Chris will fall before coming to a stop.
Recommended video:
Guided course
06:24
Conservation Of Mechanical Energy
Related Practice
Textbook Question

A vertical spring (ignore its mass), whose spring constant is 875 N/m, is attached to a table and is compressed down by 0.220 m. What upward speed can it give to a 0.380-kg ball when released?

994
views
Textbook Question

Consider the track shown in Fig. 8–39. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is a horizontal span 3.0 m long with a coefficient of kinetic friction μₖ = 0.25. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.20 m. Determine the velocity of the block at point B.

2022
views
Textbook Question

A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. If friction is present, does the skier fly off at a greater or lesser angle?

1364
views
Textbook Question

Two masses are connected by a string as shown in Fig. 8–35. Mass mA = 3.5 kg rests on a frictionless inclined plane, while mB = 5.0 kg is initially held at a height of h = 0.75 m above the floor. Use conservation of energy to find the velocity of the masses just before mB hits the floor. You should get the same answer as in part (b).

1276
views
Textbook Question

The 9.0-kg mass in Fig. 8–36 is held just barely in contact with a spring for which k = 450 N/m . When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

1909
views
Textbook Question

You slide down an 8.0-m-high icy hill (≈ frictionless). At the bottom is a level stretch where the coefficient of kinetic friction is 0.30. How far would you travel across the level stretch?

1592
views