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Chapter 7: Problem 846

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

Expert verified
(C) E / 2

Step by step solution

01

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.
02

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: \(x_Q = 2x_P\)
03

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 \(U_P\).
04

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: \(E = U_P = \frac{1}{2} k_P x_P^2\)
05

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.

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