Chapter 5, Problem 31P

A horizontal spring attached to a wall has a force constant of 850 N/m. A block of mass 1.00 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure 5.22. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched 6.00 cm from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn’t it half the speed at equilibrium?

To determine

The objects that constitute the system and through what force they interact.

Answer to Problem 31P

The objects that constitute the system are the mass, the spring and the Earth. The parts of the system will be interacting with forces normal force, gravitational force, and the spring force.

Explanation of Solution

The objects that constitute the system are the mass, the spring and the Earth. The spring and the mass will interact with the spring force. The mass and the Earth will be having an interaction force named the gravitational force due to the weight of the object. The system will be having another force named as normal force.

Thus, the objects that constitute the system are the mass, the spring and the Earth. The parts of the system will be interacting with forces normal force, gravitational force and the spring force.

Conclusion:

The objects that constitute the system are the mass, the spring and the Earth. The parts of the system will be interacting with forces normal force, gravitational force, and the spring force.

To determine

The points of interest.

Answer to Problem 31P

The points of interest are the equilibrium point and the point from which the mass is released from rest.

Explanation of Solution

There are two point of interest in this situation. One is the equilibrium point. The equilibrium point is the point where the $x=0$ point is considered. Another is the point from which the mass is released from the rest.

The point from which the mass is released from rest is $x=6.0\text{cm}$.

Thus, the points of interest are the equilibrium point and the point from which the mass is released from rest.

Conclusion:

The points of interest are the equilibrium point and the point from which the mass is released from rest.

To determine

The energy stored in the spring when the mass is stretched and when the mass is passed through equilibrium after being released from the rest.

Answer to Problem 31P

The energy stored in the spring when the mass is stretched is $1.53\text{J}$ and the energy stored in the spring when the mass is passed through equilibrium after being released from the restis $0\text{J}$.

Explanation of Solution

Section 1:

To determine: The energy stored in the spring when the mass is stretched.

Answer: The energy stored in the spring when the mass is stretched is $1.53\text{J}$.

Explanation:

Given Info:

The mass is stretched $6.0\text{cm}$ from the equilibrium.

The force constant of the spring is $850\text{N/m}$.

The mass of the object is $1.0\text{kg}$.

Formula to calculate the energy stored in the spring is,

$P{E}_s=\frac{1}{2}k{x}^2$

• $k$ is the force constant
• $x$ is the displacement of the mass from the equilibrium

Substitute $850\text{N/m}$ for $k$ and $6.0\text{cm}$ for x to find the energy stored,

$\begin{array}{c}P{E}_s=\frac{1}{2}\left(850\text{N/m}\right){\left(6.0\times {10}^{-2}\text{m}\right)}^2\\ =1.53\text{J}\end{array}$

Thus, the energy stored in the spring when the mass is stretched is $1.53\text{J}$.

Section 2:

To determine: The energy stored in the spring when the mass is passed through equilibrium after being released from the rest.

Answer: The energy stored in the spring when the mass is passed through equilibrium after being released from the restis $0\text{J}$.

Explanation:

Given Info:

At equilibrium, $x=0$.

Formula to calculate the energy stored in the spring is,

$P{E}_s=\frac{1}{2}k{x}^2$

• $k$ is the force constant
• $x$ is the displacement of the mass from the equilibrium

Substitute $850\text{N/m}$ for $k$ and $zero$ for x to find the energy stored,

$\begin{array}{c}P{E}_s=\frac{1}{2}\left(850\text{N/m}\right){\left(0\right)}^2\\ =0\text{J}\end{array}$

Thus, the energy stored in the spring when the mass is passed through equilibrium after being released from the restis $0\text{J}$.

Conclusion:

The energy stored in the spring when the mass is stretched is $1.53\text{J}$ and the energy stored in the spring when the mass is passed through equilibrium after being released from the restis $0\text{J}$.

To determine

The expression for the conservation of energy equation and the expression and numerical value of speed of the mass as it passes the equilibrium.

Answer to Problem 31P

The expression for the conservation of energy equation and the expression and numerical value of speed of the mass as it passes the equilibrium are,

$\frac{1}{2}m{v}_f{}^2+\frac{1}{2}k{x}_f{}^2=\frac{1}{2}m{v}_i{}^2+\frac{1}{2}k{x}_i{}^2$

$v_f=\sqrt{v_i{}^2+\frac{k}{m}\left(x_i-x_f\right)}$

The numerical value of the speed of the mass as it passes the equilibrium is $1.75\text{m/s}$

Explanation of Solution

Given Info:

The mass is stretched $6.0\text{cm}$ from the equilibrium.

The force constant of the spring is $850\text{N/m}$.

The mass of the object is $1.0\text{kg}$.

Since in this situation, the gravitational force and the normal force are acting in the perpendicular direction of the motion of the mass. Thus, the only force acting on the mass is the conservative spring force. Therefore the total mechanical energy of the object will be a constant.

Consider the $h=0$ point at the level of horizontal surface, according to the conservation of energy;

$K{E}_f+{\left(P{E}_g\right)}_f=K{E}_i+{\left(P{E}_s\right)}_i$

$\frac{1}{2}m{v}_f{}^2+\frac{1}{2}k{x}_f{}^2=\frac{1}{2}m{v}_i{}^2+\frac{1}{2}k{x}_i{}^2$

Formula to calculate the speed of the mass is,

$v_f=\sqrt{v_i{}^2+\frac{k}{m}\left(x_i-x_f\right)}$

• $k$ is the force constant
• $x$ is the displacement of the mass from the equilibrium

Substitute $850\text{N/m}$ for $k$, $1.0\text{kg}$ for m, $6.0\text{cm}$ for $x_i$ and $zero$ for $x_f$ and $zero$ for $v_i$ to find the final speed,

$\begin{array}{c}{v}_f=\sqrt{0+\frac{850\text{N/m}}{1.0\text{kg}}\left[{\left(6.0\times {10}^{-2}\text{m}\right)}^2-0\right]}\\ =1.75\text{m/s}\end{array}$

Conclusion:

The expression for the conservation of energy equation and the expression and numerical value of speed of the mass as it passes the equilibrium are,

$\frac{1}{2}m{v}_f{}^2+\frac{1}{2}k{x}_f{}^2=\frac{1}{2}m{v}_i{}^2+\frac{1}{2}k{x}_i{}^2$

$v_f=\sqrt{v_i{}^2+\frac{k}{m}\left(x_i-x_f\right)}$

The numerical value of the speed of the mass as it passes the equilibrium is $1.75\text{m/s}$

To determine

The speed of the mass at half way point and why is not half the speed at equilibrium.

Answer to Problem 31P

The speed of the mass at half way point is $1.51\text{m/s}$ and the speed of the mass at the half way point is not the half of the speed at equilibrium.

Explanation of Solution

Given Info:

The mass is stretched $6.0\text{cm}$ from the equilibrium.

The force constant of the spring is $850\text{N/m}$.

The mass of the object is $1.0\text{kg}$.

At the half way point $x_f=3.0\text{cm}$.

Formula to calculate the speed of the mass is,

$v_f=\sqrt{v_i{}^2+\frac{k}{m}\left(x_i-x_f\right)}$

• $k$ is the force constant
• $x$ is the displacement of the mass from the equilibrium

Substitute $850\text{N/m}$ for $k$, $1.0\text{kg}$ for m, $6.0\text{cm}$ for $x_i$ and $3.0\text{cm}$ for $x_f$ and $\text{zero}$ for $v_i$ to find the final speed,

$\begin{array}{c}{v}_f=\sqrt{0+\frac{850\text{N/m}}{1.0\text{kg}}\left[{\left(6.0\times {10}^{-2}\text{m}\right)}^2-{\left(3.0\times {10}^{-2}m\right)}^2\right]}\\ =1.51\text{m/s}\end{array}$

Thus, the speed of the mass at half way point is $1.51\text{m/s}$.

Since, the equation for final speed is not a linear equation; the speed of the mass at the half way point is not the half of the speed at equilibrium.

Conclusion:

The speed of the mass at half way point is $1.51\text{m/s}$ and the speed of the mass at the half way point is not the half of the speed at equilibrium.