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) $\mathrm{E} / 4$

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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.

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Short Answer

Step by step solution

Using Hooke's Law for both springs

Finding the relation between displacements

Setting up the potential energy equations

Relating the potential energy equations

Solving for the energy stored in spring P

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