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Chapter 5: Problem 65

\(\bullet$$\bullet\) You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{m},\) and you happen to know that your little sister weighs 220 \(\mathrm{N}\) and the wagon weighs 75 \(\mathrm{N}\) . Crossing a street, you accelerate from rest to your normal walk- ing speed of 1.5 \(\mathrm{m} / \mathrm{s}\) in 2.0 \(\mathrm{s}\) , and you notice that while you're accelerating, the bungee's length increases to about 2.0 \(\mathrm{m}\) . What's the force constant of the bungee cord, assuming it obeys Hooke's law?

Short Answer

Expert verified
Approximately 32.25 N/m.

Step by step solution

01

Identify known quantities

We need to solve for the force constant (spring constant) of the bungee cord. We are given the following:- Unstretched length of the bungee, \(L_0 = 1.3\, \text{m}\)- Stretched length of the bungee, \(L = 2.0\, \text{m}\)- Weight of the sister, \(f_{sister} = 220\, \text{N}\)- Weight of the wagon, \(f_{wagon} = 75\, \text{N}\)- Final velocity, \(v_f = 1.5\, \text{m/s}\)- Time to reach this velocity, \(t = 2.0\, \text{s}\)
02

Calculate total mass

First, calculate the total weight being pulled by the bungee, which is the sum of the sister's and the wagon's weights:\[f_{total} = f_{sister} + f_{wagon} = 220\, \text{N} + 75\, \text{N} = 295\, \text{N}\]Convert the total weight to mass (remember, weight \(=\) mass \(\times\) gravity):\[m_{total} = \frac{f_{total}}{g} = \frac{295\, \text{N}}{9.8\, \text{m/s}^2} \approx 30.1\, \text{kg}\]
03

Calculate acceleration

We know the initial velocity \(v_i = 0\, \text{m/s}\), the final velocity \(v_f = 1.5\, \text{m/s}\), and the time \(t = 2.0\, \text{s}\). Use the formula for acceleration:\[a = \frac{v_f - v_i}{t} = \frac{1.5\, \text{m/s} - 0}{2.0\, \text{s}} = 0.75\, \text{m/s}^2\]
04

Calculate force due to acceleration

Apply Newton's second law to find the force required to accelerate the mass at this rate:\[F = m_{total} \times a = 30.1\, \text{kg} \times 0.75\, \text{m/s}^2 = 22.575\, \text{N}\]
05

Calculate force constant of the bungee cord

The extension of the bungee cord is the difference between its stretched length and its unstretched length:\[\Delta L = L - L_0 = 2.0\, \text{m} - 1.3\, \text{m} = 0.7\, \text{m}\]Using Hooke's Law, \(F = k \times \Delta L\), where \(k\) is the force constant we're solving for:\[k = \frac{F}{\Delta L} = \frac{22.575\, \text{N}}{0.7\, \text{m}} \approx 32.25\, \text{N/m}\]
06

Final Result

The force constant of the bungee cord, given the described stretch and acceleration conditions, is approximately \(32.25\, \text{N/m}\).

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Key Concepts

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

Newton's second law
This fundamental principle of physics states that the force applied to an object is equal to the mass of that object multiplied by its acceleration. Newton's second law is mathematically expressed as:
  • \( F = m \cdot a \),
where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
In our exercise, the little sister and her wagon represent the object being accelerated. By using this law, we calculate the necessary force needed to bring them to a speed of 1.5 m/s in a 2-second timespan. Understanding how force is tied to mass and acceleration is pivotal when analyzing dynamic systems like a bungee cord in motion.
Hooke's law
Hooke's law describes the behavior of springs and other elastic materials. It relates the force exerted by a spring to its extension or compression from the original length. The formula according to Hooke鈥檚 law is:
  • \( F = k \cdot \Delta L \),
where \( F \) is the force exerted by the spring, \( k \) is the spring constant or force constant, and \( \Delta L \) is the change in the spring's length.
In the context of the bungee cord, the unstressed length was 1.3 m and it stretched to 2.0 m when being pulled. By using Hooke's law, we find the spring constant \( k \) by dividing the force exerted by the extension of the bungee.
Acceleration calculation
Acceleration is the rate of change of velocity of an object. When you increase your speed from rest to a walking pace, you experience acceleration. This exercise asks us to determine the acceleration by using the formula:
  • \( a = \frac{v_f - v_i}{t} \),
where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time taken to change from \( v_i \) to \( v_f \).
In this case, the wagon and sister started from rest, so \( v_i = 0 \) m/s, reached a speed of 1.5 m/s in 2 seconds. This calculation revealed the acceleration as 0.75 \( \text{m/s}^2 \). Understanding this concept is crucial when examining how quickly something changes motion.
Force constant
The force constant, or spring constant, \( k \), measures the stiffness of a spring or elastic material. A higher force constant means the material is stiffer and harder to stretch or compress.
In the bungee cord scenario, we determined the force constant by imagining the exertion needed to stretch the cord. By dividing the force (which we found using Newton鈥檚 second law) by the cord's extension (given by the difference in original and stretched lengths), we deduce the force constant of the bungee as approximately 32.25 \( \text{N/m} \). This tells us how much force is needed per meter of extension.
Bungee cord dynamics
The dynamics of a bungee cord are all about how it stretches and recoils back in response to forces, governed by laws like Newton鈥檚 and Hooke鈥檚. Bungee cords are popular in activities for their elasticity, where they must be strong enough to support weight yet elastic enough to absorb shocks and provide bounce.
  • The unstretched length gives us a baseline to measure how far the cord extends under force.
  • The force constant gives insight into the cord's stiffness.
  • An understanding of acceleration further explains how quickly a weight can shift these physical states.
By combining all these calculations, we can predict the cord's behavior under different weights and movements, which is essential for safe and thrilling bungee activities.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Stopping distance of a car. (a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by lock- ing the brakes when traveling at 29.1 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} )\) ? (b) On wet pavement, the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in ordel to be able to stop in the same distance as in part (a)? (Note. Locking the brakes is not the safest way to stop.)

\(\bullet\) Heart repair. A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a 16.0 \(\mathrm{cm}\) strip of the donated aorta reveal that it stretches 3.75 \(\mathrm{cm}\) when a 1.50 \(\mathrm{N}\) pull is exerted on it. (a) What is the force constant of this strip of aortal material? (b) If the maxi- mum distance it will be able to stretch when it replaces the aorta in the damaged heart is \(1.14 \mathrm{cm},\) what is the greatest force it will be able to exert there?

\(\bullet$$\bullet\) Mars Exploration Rover landings. In January 2004 the Mars Exploration Rover spacecraft landed on the surface of the Red Planet, where the acceleration due to gravity is 0.379 what it is on earth. The descent of this 827 kg vehicle occurred in several stages, three of which are outlined here. In Stage I, friction with the Martian atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in a 4.0 min interval. In Stage II, a parachute reduced the speed from 1600 \(\mathrm{km} / \mathrm{h}\) to 321 \(\mathrm{km} / \mathrm{h}\) in \(94 \mathrm{s},\) and in Stage III, which lasted 2.5 \(\mathrm{s}\) , retrorockets fired to reduce the speed from 321 \(\mathrm{km} / \mathrm{h}\) to zero. As part of your solu- tion to this problem, make a free-body diagram of the rocket during each stage. Assuming constant acceleration, find the force exerted on the spacecraft (a) by the atmosphere during Stage I, (b) by the parachute during Stage II, and (c) by the retrorockets during Stage III.

\(\bullet$$\bullet\) A winch is used to drag a 375 \(\mathrm{N}\) crate up a ramp at a con- stant speed of 75 \(\mathrm{cm} / \mathrm{s}\) by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at \(33^{\circ}\) above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25 . (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the accel- eration of the crate immediately after the rope broke?

\(\bullet$$\bullet\)The TGV, France's high-speed train, pulls out of the Lyons station and is accelerating uniformly to its cruising speed. Inside one of the cars, a 3.00 \(\mathrm{N}\) digital camera is hanging from the luggage compartment by a light, flexible strap that makes a \(12.0^{\circ}\) angle with the vertical. (a) Make a free-body diagram of this camera. (b) Apply Newton's second law to the camera, and find the acceleration of the train and the tension in the strap.
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