A 65.0 -kg bungee jumper steps off a bridge with a light...

A 65.0 -kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord is $11.0 \mathrm{~m}$. The jumper reaches the bottom of her motion $36.0 \mathrm{~m}$ below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 11.0 -m free fall and a 25.0 -m section of simple harmonic oscillation. (a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or nonisolated? (d) From your response in part (c) find the spring constant of the bungee cord. (e) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper? (f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by $25.0 \mathrm{~m} ?$ (h) What is the total time interval for the entire 36.0 -m drop?

A 65.0 -kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord is $11.0 \mathrm{~m}$. The jumper reaches the bottom of her motion $36.0 \mathrm{~m}$ below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 11.0 -m free fall and a 25.0 -m section of simple harmonic oscillation.
(a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or nonisolated? (d) From your response in part (c) find the spring constant of the bungee cord. (e) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper?
(f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by $25.0 \mathrm{~m} ?$
(h) What is the total time interval for the entire 36.0 -m drop?

Key Concepts

Free Fall Kinematics

This concept involves analyzing motion under a constant acceleration due to gravity when no other forces (such as air resistance) act on the object. In problems where an object is in free fall, kinematic equations—like d = (1/2) g t²—are used to determine displacement, time, or velocity. Here, the jumper’s initial 11?meter drop while the cord remains slack is an example of simple free?fall kinematics under the acceleration of gravity.

Simple Harmonic Motion (SHM)

Simple harmonic motion describes oscillatory motion where the restoring force is directly proportional to the displacement from an equilibrium position and acts in the opposite direction. In this framework, the bungee cord acts like a spring once it is taut, and its oscillatory behavior during the 25?meter stretch can be modeled using SHM equations. Key parameters include the angular frequency, given by ? = ?(k/m), and the period of oscillation, which determines the time required for specific portions of the motion.

Energy Conservation in Elastic Systems

Energy conservation in elastic systems relates the gravitational potential energy lost by the falling mass to the elastic potential energy stored in the stretched bungee cord. This principle is used to find the spring constant by equating the work done by gravity during the cord’s extension to the elastic energy stored (½ k x²). It is a crucial concept when analyzing systems where mechanical energy is transformed between gravitational and elastic forms.

Equilibrium in a Mass-Spring System

The equilibrium point in a mass-spring system is the position where the net force is zero—typically when the upward spring force exactly balances the downward gravitational force. In this case, the equilibrium extension of the bungee cord is determined by setting mg equal to kx. Recognizing this equilibrium is important for shifting the analysis of the oscillatory motion, since the motion can be treated as symmetric oscillations about this point when using SHM models.

Isolated vs Nonisolated Systems in Mechanics

This concept involves determining whether a system can be treated as isolated, meaning that no net external forces (aside from those internally canceled or conservative forces) are acting on it. In the context of the bungee jump, although gravity appears as an external force, when the system is taken to include the Earth along with the jumper and cord, gravitational forces can be regarded as internal to the combined system. This consideration is key when applying conservation laws, such as the conservation of energy, to the oscillatory (spring-like) phase of the motion.

Transcript

00:01 First range, let's start our discussion.

00:03 Suppose a bungee jumper taps off with a bridge.

00:07 Okay, and the length of the cord is 11 meter.

00:13 So up to 11 meter the length is free fall.

00:18 The body is in under free fall.

00:22 And after 11 meter, the jumper reaches to the bottom of a height of 36 meter.

00:34 That is up to 11.

00:36 Meter the motion is free fall and after 11 meter the motion is in shm okay now for the free fall part that is up to 11 meter all the linear equation of motion that is v is equal to u plus g t s is equal to u t plus half gt square v square minus u square is equal to 2 gs that is all linear motions due to under gravity will hold.

01:11 Okay, this now for what time interval is she is in free fall? okay, for what time interval bungee is in free fall? so basically we have to find out now time.

01:28 So we will use this equation as equals to ut plus half gt square.

01:33 Initially u is zero because it's take off from rest and as that up to length of 11 meter the body is under freefall gt square so 2 into 11 upon 9 .8 under root this is the value of time interval comes so t will be 1 .49 second approximately 1 .5 second up to 1 .5 second the body is under free fall now after 11 meter okay the spring and the earth is in a known isolated system because it also holds a force due to gravity.

02:20 Okay, after 25 meters the acceleration due to gravity will also act on that bungee jumper so the earth and the string system is known isolated.

02:33 Okay, now we have to find out the force constant.

02:39 See, the stretch length of the bungee jumping cord is rope or rope is 25 meters.

02:45 So force will be kx is equals to m g and k will be m g by x you can also see in this that in the force constant formula oscillation due to gravity also comes so a spring system and bungee jumping system are a non -isolated system okay so after putting the mass that 65 kg jumper 9 .8 and 25 meter the k will come to be 25 .48 newton per meter.

03:26 This is the force constant.

03:28 Now we have to find out the location of the point where spring force equally balancing the spring force balancing the gravitational force.

03:44 It is f is equal to kx equals to m g so x will be m g by minus k so it will be 65 into 9 .8 upon 25 .48 if you solve this you will get a value of minus 25 meter.

04:09 See here the minus sign only sign only sign that the applied force and the displacement acting is in different direction.

04:19 Only...

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