Chapter 6, Problem 100P

(a)

To determine

The potential energy, and kinetic energy of particle at time $t=0$.

Answer to Problem 100P

The potential energy of particle at time $t=0$ is $\underset{\_}{U=-550\text{J}}$, and kinetic energy is $\underset{\_}{K=450\text{J}}$.

Explanation of Solution

Consider the graph of potential energy plotted against distance.

It is given that at $t=0$ particle is at $x=5.5\text{cm}$, at this distance the potential energy is $U=-550\text{J}$ as shown in graph.

The total energy of the system is the sum of potential and kinetic energy. Hence, the expression for kinetic energy is given below

$K=E-U$ (I)

Here, $K$ is the kinetic energy, $E$ is the total energy, and $U$ is the potential energy

The total energy of  the system is $-100\text{J}$.

Conclusion:

Substitute $-550\text{J}$ for $U$, and $-100\text{J}$ for $E$ in equation (I)

$\begin{array}{c}K=-100\text{J}-\left(-550\text{J}\right)\\ =450\text{J}\end{array}$

Therefore, the potential energy of particle at time $t=0$ is $\underset{\_}{U=-550\text{J}}$, and kinetic energy is $\underset{\_}{K=450\text{J}}$.

(b)

To determine

The total, potential, and kinetic energy of particle at $x=1$ and moving to the right.

Answer to Problem 100P

The total energy is $\underset{\_}{E=-100\text{J}}$, potential energy is $\underset{\_}{U=-100\text{J}}$, and kinetic energy is $\underset{\_}{K=0\text{J}}$.

Explanation of Solution

The total energy of the system is already fixed, and it is $E=-100\text{J}$.

From graph potential energy at $x=1$ is $U=-100\text{J}$.

Same as part (a) use the expression for kinetic energy.

Conclusion:

Substitute $-100\text{J}$ for $U$, and $-100\text{J}$ for $E$ in equation (I) of part (a)

$\begin{array}{c}K=-100\text{J}-\left(-100\text{J}\right)\\ =0\text{J}\end{array}$

Therefore, the total energy is $\underset{\_}{E=-100\text{J}}$, potential energy is $\underset{\_}{U=-100\text{J}}$, and kinetic energy is $\underset{\_}{K=0\text{J}}$.

(c)

To determine

The kinetic energy of particle at $x=3\text{cm}$, when it moving to left.

Answer to Problem 100P

The kinetic energy of particle at $x=3\text{cm}$, when it moving to left is $\underset{\_}{K=200\text{J}}$.

Explanation of Solution

From graph the potential energy at $x=3\text{cm}$ is $-300\text{J}$.

Similar to the above cases find kinetic energy by using equation (I) in part (a)

Conclusion:

Substitute $-300\text{J}$ for $U$, and $-100\text{J}$ for $E$ in equation (I) of part (a)

$\begin{array}{c}K=-100\text{J}-\left(-300\text{J}\right)\\ =200\text{J}\end{array}$

Therefore, the kinetic energy of particle at $x=3\text{cm}$, when it moving to left is $\underset{\_}{K=200\text{J}}$.

(d)

To determine

To describe the motion of particle starting at $t=0$.

Answer to Problem 100P

The particle starting from $x=5.5\text{cm}$ is moves to both left and right with initial kinetic energy $450\text{J}$, and reaches a point where kinetic energy is zero, corresponding to $x=1\text{cm}$ left, and $x=11\text{cm}$ to the right. The cycle of motion repeats endlessly.

Explanation of Solution

At $t=0$ particle is at $x=5.5\text{cm}$, the kinetic energy of particle is $450\text{J}$. It moves towards left from this point continuously with decreasing kinetic energy until reaches at $x=1\text{cm}$. At the turning point $x=1\text{cm}$ , the kinetic energy is zero, and particle is at rest momentarily.

The particle moves to right with increasing kinetic energy till it reaches $x=5.5\text{cm}$ with kinetic energy $450\text{J}$. Kinetic energy remains same until it reaches $x=11\text{cm}$. Hereafter kinetic energy decreases by $180\text{J}$ for every centimetre until particle reaches at $x=13.5\text{cm}$. Again at this turning point the kinetic energy again become zero. Again particle turn to left and cycle of motion repeats continuously.