Chapter 6, Problem 4SP

(a)

To determine

The initial potential energy.

Answer to Problem 4SP

The initial potential energy is $3.5\text{J}$.

Explanation of Solution

Given info: The spring constant is $360\text{N/m}$ and the stretched distance is $0.14\text{m}$.

Write the expression for the elastic potential energy of a spring.

$PE=\frac{1}{2}k{x}^2$ (1)

Here,

$PE$ is the elastic potential energy

$k$ is the spring constant

$x$ is the stretched distance

Substitute $360\text{N/m}$ for $k$ and $0.14\text{m}$ for $x$ in equation (1) to find $PE$.

$\begin{array}{c}PE=\frac{1}{2}\left(360\text{N/m}\right){\left(0.14\text{m}\right)}^2\\ =3.5\text{J}\end{array}$

Conclusion:

Thus the initial potential energy of the spring-mass system is $3.5\text{J}$.

(b)

To determine

The maximum velocity that the mass will reach in its oscillation and the position where then system achieves maximum velocity.

Answer to Problem 4SP

The maximum velocity that the mass will reach in its oscillation is $5.9\text{m/s}$ and this velocity occurs as the mass moves through the equilibrium position.

Explanation of Solution

Given info: The mass attached to the spring is $0.20\text{kg}$.

The maximum velocity will be at a point where the kinetic energy of the system is maximum. The maximum kinetic energy is obtained when all the potential energy stored in the state, is completely converted to the kinetic energy. Since the initial potential energy is $3.5\text{J}$, the maximum kinetic energy that the system can achieve is also $3.5\text{J}$.

$KE=3.5\text{J}$

Write the expression for kinetic energy of an object.

$KE=\frac{1}{2}m{v}^2$

Here,

$m$ is the mass of the object

$v$ is the velocity of the object

Rewrite the above equation for $v$.

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

Substitute $3.5\text{J}$ for $KE$ and $0.20\text{kg}$ for $m$ in equation (2) to find the value of maximum $v$.

$\begin{array}{c}v=\sqrt{\frac{2\left(3.5\text{J}\right)}{0.30\text{kg}}}\\ =5.9\text{m/s}\end{array}$

The potential energy of the spring-mass system converts completely to kinetic energy when the system moves through the equilibrium position. This implies the maximum velocity occurs as the mass moves through the equilibrium position.

Conclusion:

Therefore the maximum velocity that the mass will reach in its oscillation is $5.9\text{m/s}$ and this velocity occurs as the mass moves through the equilibrium position.

(c)

To determine

The potential energy, kinetic energy and the velocity of the mass when the mass is $7\text{cm}$ from the equilibrium position.

Answer to Problem 4SP

When the mass is $7\text{cm}$ from the equilibrium position, the potential energy is $0.88\text{J}$, kinetic energy is $2.6\text{J}$ and the velocity is $5.1\text{m/s}$.

Explanation of Solution

Given info: The mass attached to spring is $0.20\text{kg}$ and the spring constant is $360\text{N/m}$.

Substitute $360\text{N/m}$ for $k$ and $7\text{cm}$ for $x$ in equation (1) to find $PE$.

$\begin{array}{c}PE=\frac{1}{2}\left(360\text{N/m}\right){\left(6\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\\ =\frac{1}{2}\left(360\text{N/m}\right){\left(0.06\text{ m}\right)}^2\\ =0.88\text{J}\end{array}$

At equilibrium position total energy is equal to the kinetic energy of the system. But as the system moves from the equilibrium position, it gains potential energy and there will be decrease in kinetic energy since total energy is constant. Hence the kinetic energy at a particular point is obtained by subtracting the potential energy at that point from the maximum kinetic energy.

Thus, the kinetic energy at the given point is obtained as,

$\begin{array}{c}KE=3.5\text{J}-0.88\text{J}\\ =2.6\text{J}\end{array}$

Substitute $2.6\text{J}$ for $KE$ and $0.20\text{kg}$ for $m$ in equation (2) to find the velocity at the given position.

$\begin{array}{c}v=\sqrt{\frac{2\left(2.6\text{J}\right)}{0.20\text{kg}}}\\ =5.1\text{m/s}\end{array}$

Conclusion:

Thus when the mass is $7\text{cm}$ from the equilibrium position, the potential energy is $0.88\text{J}$, kinetic energy is $2.6\text{J}$ and the velocity is $5.1\text{m/s}$.

(d)

To determine

The comparison of velocity of the mass at the equilibrium position to the value when it is $7\text{cm}$ away from the equilibrium position.

Answer to Problem 4SP

The velocity of the mass when it is $7\text{cm}$ away from the equilibrium position is $86\text{ \%}$ of the maximum velocity.

Explanation of Solution

Take the ratio of the values of velocity of the mass when it is $7\text{cm}$ away from the equilibrium position to when it is at equilibrium position.

$\text{ratio}=\frac{v_{7\text{ cm}}}{v_{\text{eq}}}$

Here,

$v_{7\text{ cm}}$ is the velocity of the mass when it is $7\text{cm}$ away from the equilibrium position

$v_{\text{eq}}$ is the velocity of the mass when it is at the equilibrium position

Substitute $5.1\text{m/s}$ for $v_{7\text{ cm}}$ and $5.9\text{m/s}$ for $v_{\text{eq}}$ in the above equation to determine the ratio.

$\begin{array}{c}\frac{v_{7\text{ cm}}}{v_{\text{eq}}}=\frac{5.1\text{m/s}}{5.9\text{m/s}}\\ =0.86\\ v_{7\text{ cm}}=0.86v_{\text{eq}}\end{array}$

Thus, the velocity of the mass at the position $7\text{cm}$ away from the equilibrium position is $86\%$ of that at equilibrium position.

Conclusion:

Thus the velocity of the mass when it is $7\text{cm}$ away from the equilibrium position is $86\text{ \%}$ of the maximum velocity.