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Chapter 6: Problem 38

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length.(a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

Short Answer

Expert verified
(a) 800 N; (b) Additional work: 240 J, Maximum force: 1600 N.

Step by step solution

01

Understanding the Work and Spring Force Equation

The work done on compressing the springs is given by the equation for elastic potential energy:\[ W = \frac{1}{2} k x^2 \]where \( W = 80.0 \text{ J} \) is the work done, \( k \) is the spring constant of one spring, and \( x = 0.200 \text{ m} \) is the compression distance.We need to find the spring constant \( k \). Since there are two springs in parallel, the effective spring constant is \( 2k \). The modified work equation becomes:\[ 80.0 = \frac{1}{2} \cdot 2k \cdot (0.200)^2 \]
02

Solving for the Spring Constant k

We solve the equation from Step 1 for \( k \):\[ 80.0 = k \cdot 0.040 \]\[ k = \frac{80.0}{0.040} \]\[ k = 2000 \text{ N/m} \]
03

Calculating the Force for Part (a)

The force required to hold the platform at the compressed position \( x = 0.200 \text{ m} \) is calculated by:\[ F = 2kx \]So, substituting the values, we get:\[ F = 2 \times 2000 \times 0.200 = 800 \text{ N} \].
04

Calculating Additional Work for Part (b)

To move the platform an additional 0.200 m, the total distance is 0.400 m.Initial work with 0.200 m was 80.0 J. Let's calculate the work for 0.400 m:\[ W_{0.4} = \frac{1}{2} (2k) (0.400)^2 \]\[ W_{0.4} = k \cdot 0.160 \]\[ W_{0.4} = 2000 \times 0.160 = 320.0 \text{ J} \]Additional work is:\[ W_{additional} = 320.0 - 80.0 = 240.0 \text{ J} \]
05

Calculating Maximum Force for Part (b)

Maximum force when compressed to 0.400 m is:\[ F_{max} = 2kx = 2 \times 2000 \times 0.400 = 1600 \text{ N} \].

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in a spring when it is compressed or stretched. Consider a spring that is not compressed. When you apply a force, compressing it by a distance, the spring stores energy. This energy, known as elastic potential energy, is given by:\[ E_p = \frac{1}{2} k x^2 \]- \( E_p \) is the elastic potential energy.- \( k \) is the spring constant, which measures the spring's stiffness.- \( x \) is the distance the spring is compressed or stretched from its equilibrium position.When you do 80.0 J of work compressing two springs by 0.200 meters, you're adding to their potential energy. As these springs are parallel, they experience equal compression and store energy collectively.
Spring Constant Calculation
The spring constant \( k \) is a measure of a spring's resistance to being compressed or stretched. It represents the stiffness of a spring. The higher the spring constant, the stiffer the spring.In the exercise's context, you find an effective spring constant for two parallel springs. The formula for work energy simplifies when you have both springs working together:- Total spring constant in parallel: \( 2k \)- Using the given work done: \( 80.0 = \frac{1}{2} (2k)(0.200)^2 \)Solving for \( k \), the individual spring constant becomes \( 2000 \text{ N/m} \).This calculation helps us understand how springs behave in parallel and how the spring constant impacts the force and energy calculations.
Work and Energy in Physics
Work and energy are fundamental concepts in physics. When you apply a force to an object, work is done, which changes the energy of that object. The equation for work is:\[ W = F \, \Delta x \]- \( W \) is work.- \( F \) is the force applied.- \( \Delta x \) is the distance over which the force is applied.In this exercise, you initially do 80.0 J of work compressing springs by 0.200 meters, which involves a force of 800 N. To compress them further by another 0.200 meters, you perform additional work, using total energy for 0.400 meters: \( 320.0 \text{ J} \). The additional work required is \( 240.0 \text{ J} \), resulting in a maximum force of 1600 N. Understanding these calculations shows how work and energy relate through forces and distances covered in a physical context.

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Most popular questions from this chapter

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8}\) kg (around \(150,000\) tons) and hit the ground at a speed of 12 \(\mathrm{km} / \mathrm{s}\) . (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a \(1.0-\) megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9}\) J of energy.)

CPA small block with a mass of 0.0900 \(\mathrm{kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.75). The block is originally revolving at a distance of 0.40 \(\mathrm{m}\) from the hole with a speed of 0.70 \(\mathrm{m} / \mathrm{s} .\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 \(\mathrm{m} .\) At this new distance, the speed of the block is observed to be 2.80 \(\mathrm{m} / \mathrm{s}\) . (a) What is the tension in the cord in the original situation when the block has speed \(v=0.70 \mathrm{m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation when the block has speed \(v=2.80 \mathrm{m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

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