1 Answer

1. 

   Anonymous

   2024-07-07T08:04:05-04:00Added an answer on July 7, 2024 at 8:04 am

   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 ($U_{}U\_\{\backslash text\{elastic\}\}$), 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, $m=4.00\hspace{0.17em}kgm\; =\; 4.00\; \backslash ,\; \backslash text\{kg\}$
     
   • Amplitude of oscillation, $A=0.50\hspace{0.17em}mA\; =\; 0.50\; \backslash ,\; \backslash text\{m\}$
     
   • 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 $x_{}x\_\{\backslash text\{eq\}\}$xeq​ can be found using Hooke’s law and the equilibrium condition:

   $mg=k{x}_{}mg\; =\; k\; x\_\{\backslash text\{eq\}\}$​

   At equilibrium: $x_{}=\frac{mg}{k}x\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{mg\}\{k\}$​

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

   $E=\frac{1}{2}k{A}^{2}E\; =\; \backslash frac\{1\}\{2\}kA^2$

   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 $x_{}x\_\{\backslash text\{eq\}\}$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 $U_{}=0U\_\{\backslash text\{elastic\}\}\; =\; 0$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:

   $U_g=mg(y_{}-y_{})=mg\left(2A\right)=4\cdot 9.8\cdot 2\cdot 0.5=39.2\hspace{0.17em}JU\_g\; =\; mg(y\_\{\backslash text\{max\}\}\; –\; y\_\{\backslash text\{min\}\})\; =\; mg(2A)\; =\; 4\; \backslash cdot\; 9.8\; \backslash cdot\; 2\; \backslash cdot\; 0.5\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$
   

   Total energy: $E=U_{}+K+U_g=0+0+39.2=39.2\hspace{0.17em}JE\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g\; =\; 0\; +\; 0\; +\; 39.2\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$
   

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

   At the lowest point:

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

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

   • 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: $E=U_{}+K+U_g=2k{A}^2+0+0=2\left(\frac{1}{2}k{A}^2\right)=k{A}^2=39.2\hspace{0.17em}JE\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g\; =\; 2kA^2\; +\; 0\; +\; 0\; =\; 2\; \backslash left(\; \backslash frac\{1\}\{2\}\; k\; A^2\; \backslash right)\; =\; kA^2\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$
   

   3. Equilibrium Position ( $y=0y\; =\; 0$ )

   At the equilibrium position:

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

   $x_{}=\frac{mg}{k}x\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{mg\}\{k\}$​

   Thus the elastic potential energy:

   $U_{}=\frac{1}{2}k{x}_2^{}=\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2=\frac{1}{2}\frac{m^2{g}^2}{k}U\_\{\backslash text\{elastic\}\}\; =\; \backslash frac\{1\}\{2\}\; k\; x\_\{\backslash text\{eq\}\}^2\; =\; \backslash frac\{1\}\{2\}\; k\; \backslash left(\; \backslash frac\{mg\}\{k\}\; \backslash right)^2\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{m^2\; g^2\}\{k\}$​

   • 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: $E=\frac{1}{2}k{A}^2=U_{}+K+U_{g}E\; =\; \backslash frac\{1\}\{2\}\; k\; A^2\; =\; U\_\{\backslash text\{elastic\}\}\; +\; K\; +\; U\_g$​

   Substituting:

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

   Solving for $KK$K:

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

   This gets us back to our starting total mechanical energy: $E=39.2\hspace{0.17em}JE\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$
   

   Thus, all the individual calculations should maintain the total energy value $E=39.2\hspace{0.17em}JE\; =\; 39.2\; \backslash ,\; \backslash text\{J\}$ in each state for consistency.

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