# A thrill seeking cat of mass 4.00 kg

A thrill seeking cat of mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillated vertically in SHM.

The amplitude is 0.50 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of all these three energies when the cat is

a) at its highest point

b) at its lowest point

c) at its equilibrium position.

(Exercise 14.39, 14th edition or 15th edition)

Share

1. To solve this problem, we need to analyze the energy at different points in the oscillation of the cat. We will consider elastic potential energy (${U}_{}U_\left\{\text\left\{elastic\right\}\right\}$), kinetic energy ($KK$), and gravitational potential energy (${U}_{g}U_g$). The total mechanical energy ($EE$) in simple harmonic motion is conserved and is given by the sum of these three energies.

Given:

• Mass of the cat, $m=4.00 kgm = 4.00 \, \text\left\{kg\right\}$
• Amplitude of oscillation, $A=0.50 mA = 0.50 \, \text\left\{m\right\}$
• At the highest point, the spring is unstretched, which means the spring is at its natural length.
• Spring constant, $kk$, needs to be determined.

First, let’s establish the spring constant $kk$k. At the equilibrium position, the spring is stretched by the weight of the cat. This stretch ${x}_{}x_\left\{\text\left\{eq\right\}\right\}$xeq can be found using Hooke’s law and the equilibrium condition:

$mg=k{x}_{}mg = k x_\left\{\text\left\{eq\right\}\right\}$

At equilibrium: ${x}_{}=\frac{mg}{k}x_\left\{\text\left\{eq\right\}\right\} = \frac\left\{mg\right\}\left\{k\right\}$

The total mechanical energy in the system is constant and can be expressed as:

$E=\frac{1}{2}k{A}^{2}E = \frac\left\{1\right\}\left\{2\right\}kA^2$

The three positions to analyze are:

1. Highest point (where the spring is unstretched, and the cat is at $y=Ay = A$y=A)
2. Lowest point (where the spring is most stretched, and the cat is at $y=-Ay = -A$y=A)
3. Equilibrium position (where the spring is stretched by ${x}_{}x_\left\{\text\left\{eq\right\}\right\}$xeq, and the cat is at $y=0y = 0$y=0)

### 1. Highest Point ( $y=Ay = A$ )

At the highest point:

• The spring is unstretched, so ${U}_{}=0U_\left\{\text\left\{elastic\right\}\right\} = 0$Uelastic=0
• The kinetic energy $K=0K = 0$K=0 because the velocity is zero at the turning point.
• The gravitational potential energy ${U}_{g}U_g$Ug relative to the lowest point:

${U}_{g}=mg\left({y}_{}-{y}_{}\right)=mg\left(2A\right)=4\cdot 9.8\cdot 2\cdot 0.5=39.2 JU_g = mg\left(y_\left\{\text\left\{max\right\}\right\} – y_\left\{\text\left\{min\right\}\right\}\right) = mg\left(2A\right) = 4 \cdot 9.8 \cdot 2 \cdot 0.5 = 39.2 \, \text\left\{J\right\}$

Total energy: $E={U}_{}+K+{U}_{g}=0+0+39.2=39.2 JE = U_\left\{\text\left\{elastic\right\}\right\} + K + U_g = 0 + 0 + 39.2 = 39.2 \, \text\left\{J\right\}$

### 2. Lowest Point ( $y=-Ay = -A$ )

At the lowest point:

• The spring is stretched by $2A2A$, so the elastic potential energy:

${U}_{}=\frac{1}{2}k\left(2A{\right)}^{2}=2k{A}^{2}U_\left\{\text\left\{elastic\right\}\right\} = \frac\left\{1\right\}\left\{2\right\}k\left(2A\right)^2 = 2kA^2$

• The kinetic energy $K=0K = 0$K=0 because the velocity is zero at the turning point.
• The gravitational potential energy ${U}_{g}=0U_g = 0$Ug=0 because it’s the reference point.

Total energy: $E={U}_{}+K+{U}_{g}=2k{A}^{2}+0+0=2\left(\frac{1}{2}k{A}^{2}\right)=k{A}^{2}=39.2 JE = U_\left\{\text\left\{elastic\right\}\right\} + K + U_g = 2kA^2 + 0 + 0 = 2 \left\left( \frac\left\{1\right\}\left\{2\right\} k A^2 \right\right) = kA^2 = 39.2 \, \text\left\{J\right\}$

### 3. Equilibrium Position ( $y=0y = 0$ )

At the equilibrium position:

• The spring is stretched by ${x}_{}x_\left\{\text\left\{eq\right\}\right\}$, and:

${x}_{}=\frac{mg}{k}x_\left\{\text\left\{eq\right\}\right\} = \frac\left\{mg\right\}\left\{k\right\}$

Thus the elastic potential energy:

${U}_{}=\frac{1}{2}k{x}_{2}^{}=\frac{1}{2}k{\left(\frac{mg}{k}\right)}^{2}=\frac{1}{2}\frac{{m}^{2}{g}^{2}}{k}U_\left\{\text\left\{elastic\right\}\right\} = \frac\left\{1\right\}\left\{2\right\} k x_\left\{\text\left\{eq\right\}\right\}^2 = \frac\left\{1\right\}\left\{2\right\} k \left\left( \frac\left\{mg\right\}\left\{k\right\} \right\right)^2 = \frac\left\{1\right\}\left\{2\right\} \frac\left\{m^2 g^2\right\}\left\{k\right\}$

• The kinetic energy is at its maximum because the potential energy is shared between gravitational and elastic forms.
• The gravitational potential energy:

${U}_{g}=mgAU_g = mgA$

Using conservation of mechanical energy: $E=\frac{1}{2}k{A}^{2}={U}_{}+K+{U}_{g}E = \frac\left\{1\right\}\left\{2\right\} k A^2 = U_\left\{\text\left\{elastic\right\}\right\} + K + U_g$

Substituting:

$39.2=\frac{1}{2}\frac{{m}^{2}{g}^{2}}{k}+K+mgA39.2 = \frac\left\{1\right\}\left\{2\right\} \frac\left\{m^2 g^2\right\}\left\{k\right\} + K + mgA$

Solving for $KK$K:

$K=39.2-\frac{1}{2}\frac{\left(4\cdot 9.8{\right)}^{2}}{k}-4\cdot 9.8\cdot 0.5=39.2-0.5\cdot 19.6\cdot 4-19.6K = 39.2 – \frac\left\{1\right\}\left\{2\right\} \frac\left\{\left(4 \cdot 9.8\right)^2\right\}\left\{k\right\} – 4 \cdot 9.8 \cdot 0.5 = 39.2 – 0.5 \cdot 19.6 \cdot 4 – 19.6$

This gets us back to our starting total mechanical energy: $E=39.2 JE = 39.2 \, \text\left\{J\right\}$

Thus, all the individual calculations should maintain the total energy value $E=39.2 JE = 39.2 \, \text\left\{J\right\}$ in each state for consistency.