Question

Two springs \(\mathrm{P} \& \mathrm{Q}\) of force constant \(\mathrm{k}_{\mathrm{P}} \& \mathrm{k}_{\mathrm{Q}}\left(\mathrm{k}_{\mathrm{Q}}=[\mathrm{kp} / 2]\right)\) are stretched by applying force equal magnitude. If the energy stored in \(\mathrm{Q}\) is \(\mathrm{E}\). Then what is the energy stored in \(\mathrm{P} ?\) (A) \(E\) (B) \(2 \mathrm{E}\) (C) \(\mathrm{E} / 2\) (D) \(\mathrm{E} / 4\)

Short answer

(C) E / 2

Step-by-step solution

Using Hooke's Law for both springs

Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. It can be written as: \(F = kx\) Since both springs are stretched by applying force equal magnitude, we can write the forces for both springs as: \(F_P = k_P x_P\) and \(F_Q = k_Q x_Q\) The given relation between spring constants is \(k_Q = k_P / 2\). Now we will find the relation between the displacements of the two springs.

Finding the relation between displacements

As the forces are equal, we set \(F_P = F_Q\), which gives us: \(k_P x_P = k_Q x_Q\) Using the relation \(k_Q = k_P / 2\), we can rewrite the above equation as: \(k_P x_P = (\frac{k_P}{2}) x_Q\) Upon simplifying, we get the relation between the displacements: \(x_Q = 2x_P\)

Setting up the potential energy equations

Next, we will use the potential energy formula for both springs. For spring P, the energy is: \(U_P = \frac{1}{2}k_P x_P^2\) For spring Q, the energy is: \(U_Q = \frac{1}{2}k_Q x_Q^2\) We know that the energy stored in spring Q is given as E. Thus, \(U_Q = E\). Our main goal is to find the value of \(U_P\).

Relating the potential energy equations

Since we know the relation between the displacements (as found in Step 2), we can plug this into the potential energy formulas. From \(x_Q = 2x_P\), we can get \(x_Q^2 = 4x_P^2\). Now substitute this relation and \(k_Q = k_P / 2\) into the potential energy equation for spring Q: \(E = U_Q = \frac{1}{2} (\frac{k_P}{2}) (4x_P^2)\) Simplify the equation: \(E = k_P x_P^2\) Now we can relate this to the potential energy equation for spring P: \(E = U_P = \frac{1}{2} k_P x_P^2\)

Solving for the energy stored in spring P

Now that we have the equations relating the energies stored in springs P and Q, we can solve for the energy stored in spring P (\(U_P\)). From the previous step, we have: \(E = \frac{1}{2} k_P x_P^2\) We are trying to find the value of \(U_P\), which is equal to the energy stored in spring P when the force is applied. Therefore, solving for \(U_P\), we get: \(U_P = E (\frac{1}{2})\) So, the energy stored in spring P is equal to half the energy stored in spring Q. The correct answer is (C) E / 2.

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle relating to the behavior of springs. It states that the force required to compress or extend a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it is expressed as:\[ F = kx \] where: - \( F \) is the force exerted by the spring, - \( k \) is the force constant or spring constant, - \( x \) is the displacement from the equilibrium position.In the context of the exercise, both springs \( P \) and \( Q \) are subjected to forces of the same magnitude. This translates to equal force expressions for each spring: \( F_P = k_P x_P \) and \( F_Q = k_Q x_Q \). Understanding this is crucial as it allows us to explore how differences in spring constants affect displacements as seen in the relationship \( x_Q = 2x_P \).

Potential Energy in Springs

Potential energy in springs, often called elastic potential energy, refers to the energy stored as a spring is compressed or stretched. The amount of potential energy stored in a spring can be calculated with:\[ U = \frac{1}{2}kx^2 \]where:- \( U \) is the potential energy stored in the spring,- \( k \) is the spring constant,- \( x \) is the displacement from the equilibrium position.In our exercise, spring \( Q \) stores a potential energy of \( E \), calculated as \( \frac{1}{2}k_Q x_Q^2 \). By substituting this into the expression for spring \( P \), and using the relationship \( x_Q = 2x_P \), we establish that the energy in spring \( P \) is half of that in spring \( Q \) due to the shared force and varied displacement. This results in \( U_P = \frac{E}{2} \).

Force Constant

The force constant, or spring constant, denoted as \( k \), is a measure of the stiffness of a spring. A larger \( k \) value indicates a stiffer spring which requires more force to achieve a certain displacement. The relevance of the force constant in our exercise is highlighted with the values for springs \( P \) and \( Q \), where \( k_Q = \frac{k_P}{2} \).This differing stiffness impacts the displacement observed when both springs are subjected to equal forces. With \( k_Q \) being half of \( k_P \), spring \( Q \) deflects more to accommodate the same force. This is captured by the displacement relationship \( x_Q = 2x_P \), illustrating how a lower force constant leads to greater flexibility and thus, more displacement. Understanding force constants helps explain the variance in energy storage between the springs.