Chapter 7, Problem 66CP

A particle of mass m = 1.18 kg is attached between two identical springs on a frictionless, horizontal tabletop. Both springs have spring constant k and are initially unstressed, and the particle is at x = 0. (a) The particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs as shown in Figure P7.50. Show that the force exerted by the springs on the particle is

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

(b) Show that the potential energy of the system is

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

(c) Make a plot of U(x) versus x and identify all equilibrium points. Assume L = 1.20 m and k = 40.0 N/m. (d) If the panicle is pulled 0.500 m to the right and then released, what is its speed when it reaches x = 0?

Figure P7.50

To determine

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

Answer to Problem 66CP

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

Explanation of Solution

Given info: The mass of the particle is $1.18\text{kg}$, the spring constant of both the springs is $k$.

The free body diagram of the given case is as shown in the figure below.

Figure (1)

The extension in the spring is,

$\Delta x=x_2-L$ (1)

Here,

$x_2$ is the length after stretching.

$L$ is the initial stretch.

The new length after stretching is,

$x_2=\sqrt{x^2+L^2}$

Here,

$x$ is the distance the spring is pulled.

Substitute $\sqrt{x^2+L^2}$ for $x_2$ in equation (1).

$\Delta x=\sqrt{x^2+L^2}-L$

From the figure (1) the net force in $y$ direction is zero.

The net force in $x$ direction is,

${\overrightarrow{\text{F}}}_{\text{kx}}=-2F_{\text{k}}\cos \theta \widehat{\text{i}}$ (2)

The negative sign is due to the direction of the force in the negative direction.

Here,

$F_{\text{k}}$ is the spring force.

$\theta$ is the angle at which spring is pulled.

From the free body diagram the value of $\cos \theta$ is,

$\cos \theta =\frac{x}{\sqrt{x^2+L^2}}$

The formula for the spring force is,

$F_{\text{k}}=k\Delta x$

Substitute $k\Delta x$ for $F_{\text{k}}$ and $\frac{x}{\sqrt{x^2+L^2}}$ for $\cos \theta$ in the equation (2).

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

Substitute $\sqrt{x^2+L^2}-L$ for $\Delta x$ in the above equation.

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

Write the expression for the force exerted by the spring on the particle.

$\overrightarrow{\text{F}}={\overrightarrow{\text{F}}}_{\text{kx}}+{\overrightarrow{\text{F}}}_{\text{ky}}$

Substitute $-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{F}}}_{\text{kx}}$ and $0$ for ${\overrightarrow{\text{F}}}_{\text{ky}}$ in above equation.

$\begin{array}{c}\overrightarrow{F}=\left(-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\right)\widehat{\text{i}}+\left(0\right)\widehat{\text{j}}\\ \text{=}\left(-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\right)\widehat{\text{i}}\end{array}$

Conclusion:

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

To determine

To show: The 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 66CP

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

Explanation of Solution

Given info: The mass of the particle is $1.18\text{kg}$, the spring constant of both the springs is $k$.

From part (a) the force exerted by the spring on the particle is,

$F_{kx}=-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)$

The potential energy of a system is,

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

Substitute $-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)$ for $F_{\text{kx}}$ in the above equation.

$\begin{array}{c}U\left(x\right)=-{\displaystyle {\int }_0^x\left(-2kx\left(1-\frac{L}{\sqrt{x^2+L^2}}\right)\right)dx}\\ =2k{\displaystyle {\int }_0^{x}xdx}-2kL{\displaystyle {\int }_0^x\frac{x}{\sqrt{x^2+L^2}}}dx\\ =k{x}^2+2kL\left(L-\sqrt{x^2+L^2}\right)\end{array}$

Conclusion:

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

To determine

To draw: The plot of $U\left(x\right)$ versus $x$ and equilibrium points from the graph.

Answer to Problem 66CP

The plot for $U\left(x\right)$ versus $x$ is shown below and at $x=0$ is the only equilibrium point.

Explanation of Solution

Introduction:

The equilibrium points are the points at which the values of the force and the potential energy have the minimum value in order to have higher stability in the system.

Given info: The mass of the particle is $1.18\text{kg}$, the spring constant of both the springs is $k$, the length of spring is $1.20\text{m}$ and the spring constant is $40.0\text{N}/\text{m}$.

The potential energy of the system is,

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

Substitute $40.0\text{N}/\text{m}$ for $k$, $1.20\text{m}$ for $L$ in the above equation.

$\begin{array}{c}U\left(x\right)=\left(40.0\text{N}/\text{m}\right)x^2+2\left(40.0\text{N}/\text{m}\right)\left(1.20\text{m}\right)\left(\left(1.20\text{m}\right)-\sqrt{x^2+{\left(1.20\text{m}\right)}^2}\right)\\ =\left(40.0\text{N}/\text{m}\right)x^2+96.0\text{N}\left(1.20\text{m}-\sqrt{x^2+1.44{\text{m}}^2}\right)\end{array}$

Thus, the potential energy of the system is,

$U\left(x\right)=\left(40.0\text{N}/\text{m}\right)x^2+96.0\text{N}\left(1.20\text{m}-\sqrt{x^2+1.44{\text{m}}^2}\right)$

The plot $U\left(x\right)$ versus $x$ for $-1.0<x<1.0$ is as shown in the figure below.

Figure (2)

The equilibrium points are the point at the potential energy is zero. In the above plot the minimum potential energy is at $x=0$ so this points are considered to be equilibrium point.

Form the graph the equilibrium point is for $x=0$ for the given case.

Conclusion:

Therefore, the plot for $U\left(x\right)$ versus $x$ is shown in figure (2) and the equilibrium point is $x=0$.

To determine

The speed of the particle.

Answer to Problem 66CP

The speed of the particle is $0.823\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the particle is $1.18\text{kg}$, the spring constant of both the springs is $k$, the length of spring is $1.20\text{m}$, the spring constant is $40.0\text{N}/\text{m}$ and the length the spring is pulled is $0.500\text{m}$.

The potential energy of the system is,

$U\left(x\right)=\left(40.0\text{N}/\text{m}\right)x^2+96.0\text{N}\left(1.20\text{m}-\sqrt{x^2+1.44{\text{m}}^2}\right)$

Substitute $0.500\text{m}$ for $x$ in the above equation.

$\begin{array}{c}U\left(x\right)=\left(40.0\text{N}/\text{m}\right){\left(0.500\text{m}\right)}^2+96.0\text{N}\left(1.20\text{m}-\sqrt{{\left(0.500\text{m}\right)}^2+1.44{\text{m}}^2}\right)\\ =0.4\text{J}\end{array}$

Thus, the potential energy of the system is $0.4\text{J}$.

The potential energy is converted in to the kinetic energy to follow the law of conservation of momentum.

$U\left(x\right)=\frac{1}{2}m{v}^2$

Here,

$v$ is the velocity of the of the particle.

Rearrange the above equation for $v$.

$v=\sqrt{\frac{2U\left(x\right)}{m}}$

Substitute $1.18\text{kg}$ for $m$ and $0.4\text{J}$ for $U\left(x\right)$ in the above equation.

$\begin{array}{c}v=\sqrt{\frac{2\left(0.4\text{J}\right)}{1.18\text{kg}}}\\ =0.823\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, speed of the particle is $0.823\text{m}/\text{s}$.