Chapter 6, Problem 73P

(a)

To determine

The x -component of the force as a function of x.

Explanation of Solution

Given:

The mass of the block is $m$ . The vertical distance of the top of the block and the ceiling is $y_0$ and the horizontal displacement is $x$ . Spring constant is $k$ .

Formula used:

Write the expression for the force exerted by a spring to a particle.

  $F=kx$

Here, $F$ is the force applied, $k$ is spring constant and $x$ is the displacement of the particle at any instant of time.

Calculation:

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

  $F=kz$   ........ (1)

Here, $z$ is the distance by which the spring is stretched.

Write the expression for $z$ .

  $z\text{}=\text{}l-y_0$   ........ (2)

Here, $l$ is the length of the spring.

Write the expression for $l$ .

  $l=\sqrt{\left(x^2+y_0^2\right)}$

Substitute $\sqrt{\left(x^2+y_0^2\right)}$ for $l$ in expression (2).

  $z=\sqrt{\left(x^2\text{}+\text{}{y}_0^2\right)}-y_0$

Substitute $\left(\sqrt{\left(x^2+y_0^2\right)}-y_0\right)$ for z in expression (1).

  $F=k\left(\sqrt{\left(x^2+y_0^2\right)}-y_0\right)$

Write the expression for the $x-$ component of the force.

  $F_x=F\cos \theta$   ........ (3)

Substitute $\frac{x}{\sqrt{\left(x^2+y_0^2\right)}}$ for $\cos \theta$ in expression (3).

  $F_x=\frac{F\text{}x}{\sqrt{\left(x^2+y_0^2\right)}}$   ........ (4)

Substitute $k\left(\sqrt{\left(x^2+y_0^2\right)}-y_0\right)$ for $F$ in expression (4).

  $\begin{array}{c}{F}_x=\frac{F\text{}x}{\sqrt{\left(x^2+y_0^2\right)}}\\ =\frac{k\left(\sqrt{\left(x^2+y_0^2\right)}-y_0\right)x}{\sqrt{\left(x^2+y_0^2\right)}}\\ =kx\left(1-\frac{y_0}{\sqrt{\left(x^2+y_0^2\right)}}\right)\end{array}$   ........(5)

Conclusion:

Thus, the x-component of the force is $kx\left(1-\frac{y_0}{\sqrt{\left(x^2+y_0^2\right)}}\right)$.

(b)

To determine

Whether the $x-$ component of the force is proportional to $x^3$ for sufficiently small values of $\left|x\right|$ .

Explanation of Solution

Calculation:

Rearrange the expression (5).

  $\begin{array}{c}{F}_x=kx\left(1-\frac{1}{\sqrt{\left(1+{\left(\frac{x}{y_0}\right)}^2\right)}}\right)\\ =kx\left(1-{\left(1+{\left(\frac{x}{y_0}\right)}^2\right)}^{-1/2}\right)\end{array}$

Expand the above expression in Taylor’s series and approximate up to second-order term.

  $\begin{array}{c}{F}_x\simeq kx\left(1-1-\frac{1}{2}{\left(\frac{x}{y_0}\right)}^2\right)\\ =kx\left(-\frac{1}{2}\frac{x^2}{y_0^2}\right)\\ =-\frac{k{x}^3}{2y_0^2}\end{array}$

Conclusion:

Thus, the $x-$ component of the force is proportional to $x^3$ for sufficiently small values of $\left|x\right|$ .

(c)

To determine

The speed of the block.

Explanation of Solution

Given:

The block is released from rest at $x=x_0$ , where $\left|x\right|<<y_0$ .

Formula Used:

Write the expression for the relation between final velocity, initial velocity, acceleration and the displacement of a particle.

  $v^2=u^2+2aS$

Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $S$ is the displacement of the particle.

Calculation:

Write the expression for the acceleration of the particle.

  $a=\frac{F_x}{m}$

Substitute $\frac{k{x}^3}{2y_0^2}$ for $F_x$ in the above expression.

  $a=\frac{k{x}^3}{2m{y}_0^2}$

Write the expression for the velocity of the particle when it crosses $x=0$ starting from $x=x_0$ at rest.

  $v=2a{x}_0$

Substitute $\frac{k{x}_0^3}{2m{y}_0^2}$ for $a$ in the above expression.

  $v=\frac{k{x}_0^4}{m{y}_0^2}$

Conclusion:

Thus, the speed of the block is $\frac{k{x}_0^4}{m{y}_0^2}$ .