Chapter 6, Problem 68P

(a)

To determine

To show force exerted by the springs on the particle is $\overrightarrow{\text{F}}=-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}$.

Answer to Problem 68P

The force exerted by the springs on the particle is shown as $\underset{\_}{\overrightarrow{\text{F}}=-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}}$.

Explanation of Solution

Figure 1 represents two springs attached to mass $m$.

Write the expression for force.

    $F=-kx$        (I)

Here, $F$ is the force, $k$ is the spring constant, and $x$ is the extension length.

From Figure 1 the new length is given as $\sqrt{x^2+L^2}\cos \theta$. The new extension is new length minus the original length $x=\left(\sqrt{x^2+L^2}-L\right)\cos \theta$.

Apply the above condition in equation (I).

    $F=-k\left(\sqrt{x^2+L^2}-L\right)\cos \theta$        (II)

From Figure 1, the angle cosine of $\theta$ is given as

  $\text{cos}\theta \text{=}\frac{x}{\sqrt{x^2+L^2}}$        (III)

Use equation (III) in (II).

    $\begin{array}{c}F=-k\left(\sqrt{x^2+L^2}-L\right)\frac{x}{\sqrt{x^2+L^2}}\widehat{\text{i}}\\ =-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}\end{array}$        (IV)

Conclusion:

Therefore, the force exerted by the springs on the particle is shown as $\underset{\_}{\overrightarrow{\text{F}}=-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}}$.

(b)

To determine

To show potential energy of the system is $U\left(x\right)=k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)$.

Answer to Problem 68P

The potential energy of the system is shown as $\underset{\_}{U\left(x\right)=k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)}$.

Explanation of Solution

From subpart (a) force is

    $F=-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}$

Write the potential energy of the system.

    $U\left(x\right)=-{\displaystyle {\int }_0^x{F}_{\text{x}}dx}$        (V)

Here, $U\left(x\right)$ is the potential energy, and $F_{\text{x}}$ is the force along $x$ direction.

Use equation (IV) in (V) and Integrate Equation (V) from $0$ to $x$.

    $\begin{array}{c}U\left(x\right)=-{\displaystyle {\int }_0^{\text{x}}\left(-2kx+\frac{2kLx}{\sqrt{x^2+L^2}}\right)}dx\\ =2k{\displaystyle {\int }_0^{x}xdx}-2kL{\displaystyle {\int }_0^x\frac{x}{x^2+L^2}}dx\\ =k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)\end{array}$

Conclusion:

Therefore, the potential energy of the system is shown as $\underset{\_}{U\left(x\right)=k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)}$.

(c)

To determine

To make a plot between $U\left(x\right)$ and $x$ identifying all the equilibrium points.

Answer to Problem 68P

The plot between $U\left(x\right)$ and $x$ is plotted.

Explanation of Solution

Figure 1 represents the plot between $U\left(x\right)$ and $x$.

Conclusion:

Therefore, The plot between $U\left(x\right)$ and $x$ is plotted.

(d)

To determine

The speed of the mass $m$ when it reaches $x=0$.

Answer to Problem 68P

The speed of the mass $m$ when it reaches $x=0$ is $\underset{\_}{0.823\text{m}/\text{s}}$.

Explanation of Solution

From subpart (b) gravitational potential energy is given as

    $\Delta U=k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)$        (VI)

The speed of the particle is equal to the change in gravitational potential energy.

    $\frac{1}{2}m{v}^2=-\Delta U$        (VII)

Rearrange equation (VI), to find $v$.

    $v=\sqrt{\frac{-2\Delta U}{m}}$        (VIII)

Conclusion:

Substitute $40.0\text{N}/\text{m}$ for $k$, $1.20\text{m}$ for $L$, and $0.500\text{m}$ for $x$ in equation (VI), to find $\Delta U$.

    $\begin{array}{c}\Delta U=\left(40.0\text{N}/\text{m}\right){\left(0.500\text{m}\right)}^2+2\left(40.0\text{N}/\text{m}\right)\left(0.500\text{m}\right)\left(1.20\text{m}-\sqrt{{\left(0.500\text{m}\right)}^2+{\left(1.20\text{m}\right)}^2}\right)\\ =0.400\text{J}\end{array}$

Substitute $0.400\text{J}$ for $\Delta U$, and $1.18\text{kg}$ for $m$ in equation (VIII, to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{-2\left(0.400\text{J}\right)}{\left(1.18\text{kg}\right)}}\\ =0.823\text{m}/\text{s}\end{array}$

Therefore, the speed of the mass $m$ when it reaches $x=0$ is $\underset{\_}{0.823\text{m}/\text{s}}$.