Problem 40 - University Physics with Modern Physics

Question

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 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 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 Problem

You have two parallel springs and compress them together, which requires 80.0 J of work. You need to find the force required to hold the compression at 0.200 m and the additional work to compress it another 0.200 m, as well as the force needed for this second compression.

Step 2: Relate Work to Spring Force

According to Hooke's Law, the force applied to a spring is proportional to its compression, given by the formula \(F = kx\), where \(k\) is the spring constant and \(x\) is the compression distance. The work done on the spring is given by \(W = \frac{1}{2}kx^2\).

Step 3: Calculate Spring Constant (k)

Using the work-energy principle, where work done is 80.0 J and compression is 0.200 m, calculate the effective spring constant for both springs combined using \(W = \frac{1}{2}kx^2\). Solving, \(80.0 = \frac{1}{2}k(0.200)^2\), results in \(k = 4000\, \text{N/m}\). This is the combined spring constant for the two springs.

Step 4: Calculate Force for First Compression

To find the force required to hold the platform at 0.200 m compression, use Hooke's Law \(F = kx\). Substituting the values found, \(F = 4000 \times 0.200 = 800\, \text{N}\). This is the force needed for initial compression.

Step 5: Calculate Additional Work for Further Compression

For additional 0.200 m compression, total compression distance is 0.400 m. Calculate new work done using \(W = \frac{1}{2}kx^2\) for 0.400 m, so \(W = \frac{1}{2}4000(0.400)^2 = 320.0\, \text{J}\). The additional work is the difference, \(320.0 - 80.0 = 240.0\, \text{J}\).

Step 6: Calculate Maximum Force for Further Compression

Using Hooke's Law again for total 0.400 m compression, \(F = 4000 \times 0.400 = 1600\, \text{N}\). This is the maximum force required for further compression.

Key Concepts

Spring Constant

The spring constant is a key part of understanding how springs behave when compressed or stretched. It is commonly denoted by the symbol \(k\) and represents the stiffness of a spring. A larger spring constant indicates a stiffer spring, requiring more force to compress or stretch it by a particular distance.

In our workout example, we have two springs acting together in a parallel arrangement. To determine the spring constant of both springs combined, we use the work-energy principle.

The work done on the springs is given by the equation \(W = \frac{1}{2}kx^2\). Given the work \(W = 80.0 \text{ J}\) and compression \(x = 0.200 \text{ m}\), we can solve for the combined spring constant. The calculation goes like this:\[80.0 = \frac{1}{2}k(0.200)^2\]Solving this yields \(k = 4000 \text{ N/m}\). So, the combined spring constant for these springs is 4000 N/m. This implies the springs are quite stiff, suitable for resisting significant forces.

Understanding the spring constant allows you to calculate the force and energy involved in compressing or stretching the spring, a crucial step for analyzing mechanical systems involving springs.

Work-Energy Principle

The work-energy principle is an essential concept that connects the work done on an object to its energy changes. It's essential for problems involving springs, as it helps us understand the relationship between the force applied, the distance moved, and the energy stored or used.

In the exercise, you perform 80.0 J of work to compress the springs by 0.200 meters. The principle here is that the work done on an object is stored as potential energy in the spring due to its compression or extension. Using the formula \(W = \frac{1}{2}kx^2\), we calculate how much work is needed to hold or further compress the springs.

When further compressing the springs by an additional 0.200 meters, we need to calculate the new work done for a total compression of 0.400 meters. This is calculated as:\[W = \frac{1}{2}4000(0.400)^2 = 320.0 \text{ J}\]The additional work required is the difference between the total work for 0.400 m and the initial 0.200 m compression: \(320.0 \text{ J} - 80.0 \text{ J} = 240.0 \text{ J}\).

This principle elegantly accounts for how energy is transferred and stored, allowing for detailed analysis of mechanical systems.

Force Calculation

Calculating the force needed to compress or stretch the springs involves Hooke's Law, which states that the force \(F\) required to compress or extend a spring is directly proportional to the distance \(x\) it is compressed or extended. The equation is \(F = kx\), where \(k\) is the spring constant.

For the initial compression of 0.200 meters, we calculate as follows: \(F = 4000 \times 0.200 = 800 \, \text{N}\). This is the force needed to hold the compression initially.

For further compression to 0.400 meters, the new force calculation will be: \(F = 4000 \times 0.400 = 1600 \, \text{N}\). This is the maximum force you must apply for the additional compression.

These calculations are crucial in mechanical design and analysis because they tell us precisely how much effort is needed to manipulate springs in systems ranging from simple exercise equipment to complex engineering machinery. By knowing the forces involved, you can ensure systems are designed safely and efficiently.