Problem 38 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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:\[ 80.0 = \frac{1}{2} \cdot 2k \cdot (0.200)^2 \]

Step 2: Solving for the Spring Constant k

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

Step 3: 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:\[ F = 2 \times 2000 \times 0.200 = 800 \text{ N} \].

Step 4: 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:\[ W_{additional} = 320.0 - 80.0 = 240.0 \text{ J} \]

Step 5: Calculating Maximum Force for Part (b)

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

Key Concepts

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.