Chapter 13, Problem 15P

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 47.0 J and a maximum displacement from equilibrium of 0.240 m. (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is 3.45 m/s, what is its mass? (d) What is the speed of the block when its displacement is 0.160 m? (e) Find the kinetic energy of the block at x = 0.160 m. (f) Find the potential energy stored in the spring when x = 0.160 m. (g) Suppose the same system is released from rest at x = 0.240 m on a rough surface so that it loses 14.0 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?

To determine

The spring constant of a spring.

Answer to Problem 15P

The spring constant is $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$.

Explanation of Solution

Given info: The total mechanical energy of the block-spring system is 47.0 J. The maximum displacement from equilibrium is 0.240 m.

Explanation:

The total energy of the object-spring system at any instance is given by:

$E=KE+P{E}_s$

Here,

$E$ is the total energy of the system

$KE$ is the kinetic energy of the system

$P{E}_s$ is the elastic potential energy of the system

The kinetic energy of the system is given by:

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

Here,

$m$ is the mass of the system

$v$ is the velocity of the system

The elastic potential of the system is given by:

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

Here,

$k$ is the force constant of the spring

$x$ is the displacement of the spring from its equilibrium position

At the maximum displacement point the total energy is stored as elastic potential energy.

Hence

$\frac{1}{2}k{x}_{\max }^2=E_{tot}$

Here,

$k$ is the spring constant of the spring

$x_{\max }^{}$ is the maximum displacement from the equilibrium or the amplitude

$E_{tot}$ is the total energy

On re-arrangement

$k=\frac{2E_{tot}}{x_{\max }^2}$

Substituting $E_{tot}$ as 47.0 J, $x_{\max }^{}$ as 0.240 m to find the spring constant.

$\begin{array}{l}k=\frac{2(47.0\text{J})}{{(0.240\text{m)}}^2}\\ =1.63\times {10}^3{\text{Nm}}^{\text{-1}}\end{array}$

Conclusion: The spring constant is $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$.

To determine

The kinetic energy of the system at equilibrium position.

Answer to Problem 15P

The kinetic energy of the system is 47.0 J.

Explanation of Solution

Given info: The total mechanical energy of the block-spring system is 47.0 J. The maximum displacement from equilibrium is 0.240 m.

Explanation:

The total energy of the object-spring system at any instance is given by:

The kinetic energy of the system is given by:

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

The elastic potential of the system is given by:

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

In the equilibrium position, the speed of the block is maximum and since the displacement $x$ is zero, the entire energy is stored as kinetic energy.

Hence,

$KE=E_{tot}=47.0\text{J}$

Conclusion: The kinetic energy at the equilibrium position is 47.0 J.

To determine

The mass of the block.

Answer to Problem 15P

The mass of the block is 7.90 Kg.

Explanation of Solution

Given info: The total mechanical energy of the block-spring system is 47.0 J. The maximum displacement from equilibrium is 0.240 m. The maximum speed of the block is $3.45{\text{ms}}^{\text{-1}}$.

Explanation:

The total energy of the object-spring system at any instance is given by:

The kinetic energy of the system is given by:

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

The elastic potential of the system is given by:

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

In the equilibrium position, the speed of the block is maximum and since the displacement $x$ is zero, the entire energy is stored as kinetic energy.

Hence

$\begin{array}{l}K{E}_{\max }=E_{tot}\\ \frac{1}{2}m{v}_{\max }^2=E_{tot}\\ \end{array}$

On re-arranging

$m=\frac{2E_{tot}}{v_{\max }^2}$

Substituting $E_{tot}$ as 47.0 J, $v_{\max }$ as $3.45{\text{ms}}^{\text{-1}}$ to find the mass $m$

$\begin{array}{l}m=\frac{2(47.0\text{J})}{{(3.45{\text{ms}}^{\text{-1}})}^2}\\ =7.90\text{Kg}\end{array}$

Conclusion: The mass of the block is 7.90 Kg.

To determine

The speed of the block.

Answer to Problem 15P

The speed of the block is $2.57{\text{ms}}^{\text{-1}}$.

Explanation of Solution

Given info: The total mechanical energy of the block-spring system is 47.0 J. The maximum displacement from equilibrium is 0.240 m. The maximum speed of the block is $3.45{\text{ms}}^{\text{-1}}$.

Explanation:

The total energy of the object-spring system at any instance is given by:

The kinetic energy of the system is given by:

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

The elastic potential of the system is given by:

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

Hence the total energy

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

On re-arranging

$v=\sqrt{\frac{2E-k{x}^2}{m}}$

Substituting E as 47.0 J, $k$ as $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$, $x$ as 0.160 m,  m as 7.90 Kg to find the velocity:

$\begin{array}{l}v=\sqrt{\frac{2(47.0J)-(1.63\times {10}^3{\text{Nm}}^{\text{-1}}){(0.160\text{m})}^2}{7.90\text{Kg}}}\\ =2.57{\text{ms}}^{\text{-1}}\end{array}$

Conclusion: The velocity of the block is $2.57{\text{ms}}^{\text{-1}}$.

To determine

The kinetic energy of the block.

Answer to Problem 15P

The kinetic energy is 26.7 J.

Explanation of Solution

Given info: The total mechanical energy of the block-spring system is 47.0 J. The maximum displacement from equilibrium is 0.240 m. The displacement of the block from the equilibrium position is 0.160 m. The velocity of the block is $2.57{\text{ms}}^{\text{-1}}$. The mass of the block is 7.90 Kg.

Explanation:

The kinetic energy of the system is given by:

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

From the conclusion of the previous section, the velocity of the block is $2.57{\text{ms}}^{\text{-1}}$, the mass of the block is 7.90 Kg.

Substituting 7.90 Kg for $m$, $2.57{\text{ms}}^{\text{-1}}$ for $v$ to find out the kinetic energy:

$\begin{array}{l}KE=\frac{1}{2}(7.90\text{Kg}){(}^{2.57}\\ =26.7\text{J}\end{array}$

Conclusion: The kinetic energy of the block is 26.7 J.

To determine

The potential energy stored in the spring.

Answer to Problem 15P

The potential energy stored is 20.9 J.

Explanation of Solution

Given info: The displacement of the spring from its equilibrium position is 0.160 m.

Explanation:

The elastic potential of the system is given by:

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

From the conclusion of the section (a), the spring constant is $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$.

Substituting $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$ for $k$, 0.160 for $x$ to find the potential energy stored:

$\begin{array}{l}PE=\frac{1}{2}(1.63\times {10}^3{\text{Nm}}^{\text{-1}}){(}^{0.160}\\ =20.9\text{J}\end{array}$

Conclusion: The potential energy stored in the spring when the displacement of the spring from the equilibrium position is 0.160 m is 20.9 J.

To determine

The position of the first turning point of the block.

Answer to Problem 15P

The first turning point of the block is $-0.201\text{m}$.

Explanation of Solution

Given info: The system is released from rest at x=0.240 m, by the time it reaches its first turning point it loses 14.0 J of its energy.

Explanation:

The total energy of the object-spring system at any instance is given by:

$E=KE+P{E}_s$

The kinetic energy of the system is given by:

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

The elastic potential of the system is given by:

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

Consider the right side of the equilibrium position to be positive. At the turning point, the velocity of the object briefly becomes 0. Therefore, the entire energy available is stored as elastic potential energy.

The available energy will be

$E_{avl}=\frac{1}{2}k{x}_{\max }^2-E_{lost}$

Substitute $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$ for k, 0.240 m for $x_{\max }^{}$, 14.0 J for $E_{lost}$ to find $E_{avl}$:

$\begin{array}{l}{E}_{avl}=\frac{1}{2}(1.63\times {10}^3{\text{Nm}}^{\text{-1}}){(}^{0.240}-14.0\text{J}\\ =33\text{J}\end{array}$

Now at the turning point the available energy will be stored in elastic potential energy.

$\frac{1}{2}k{x}^2=E_{avl}$

On re-arranging,

$x=\sqrt{\frac{2E_{avl}}{k}}$

Substituting 33 J for $E_{avl}$ and $1.63\times {10}^3{\text{Nm}}^{\text{-1}}$ for $k$ to find out x.

$\begin{array}{l}x=\sqrt{\frac{2(33\text{J})}{1.63\times {10}^3{\text{Nm}}^{\text{-1}}}}\\ =0.201\text{m}\end{array}$

Conclusion: The position at the first turning point is $-0.201\text{m}$.