# A cart attached to a spring vibrates with amplitude A. (a) What f

A cart attached to a spring vibrates with amplitude A.
(a) What fraction of the total energy of the cart-spring system is elastic potential energy and what fraction is kinetic energy when the cart is at position $x=\frac{A}{2}?$
(b) At what position is the cart when its kinetic energy equals its elastic potential energy?
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Bubich13
Step 1
We have a cart that attached to a spring vibrates with amplitude A. And we would like to:
a) Find out what fraction of the total energy of the system is clastic potential energy and what fraction is kinetic energy when the cart is at position $x=\frac{A}{2}$.
b) Find the position of the cart when its kinetic energy equals the elastic potential energy.
Step 2
a) The total energy of the system could be described as follows:
$E=U+K$
Where $U=\frac{1}{2}k{x}^{2}$
$K=\frac{1}{2}m{v}^{2}$
A critical cases we can use is:
$E={U}_{max}=\frac{1}{2}k{A}^{2}$
Now, let's find ${U}_{s}$, when $x=\frac{A}{2}$
${U}_{s}=\frac{1}{2}k{\left(\frac{A}{2}\right)}^{2}$
${U}_{s}=\frac{1}{4}\left(\frac{1}{2}k{A}^{2}\right)$
Notice that $E=\frac{1}{2}k{A}^{2}$. Hence ${U}_{s}=\frac{1}{4}E$
$\frac{{U}_{s}}{E}=\frac{1}{4}$
Step 3
In order to find the fraction of the total energy that is kinetic energy, we need to subtract the potential energy from the total energy to get the kinetic energy of the system at $x=\frac{A}{2}$, after that, we can find the fraction:
$K=E-{U}_{s}=\left(\frac{1}{2}k{A}^{2}\right)-\frac{1}{4}\left(\frac{1}{2}k{A}^{2}\right)$
$K=E-{U}_{s}=\frac{3}{8}k{A}^{2}$
$\frac{K}{E}=\frac{\frac{3k{A}^{2}}{8}}{\frac{k{A}^{2}}{2}}$
$\frac{K}{E}=\frac{3}{4}$
Step 4
b) When the kinetic energy equals the elastic potential energy this means that each one of this energies will be having a value of $\frac{E}{2}$:
${U}_{s}=K$
$\frac{1}{2}k{x}^{2}=\frac{E}{2}$
Notice that $E={U}_{max}=\frac{1}{2}k{A}^{2}$. Hence
$\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(\frac{1}{2}k{A}^{2}\right)$
${x}^{2}=\frac{1}{2}{A}^{2}$
$x=\sqrt{\frac{1}{2}}A$

enhebrevz
Total Energy $=\frac{1}{2}K{A}^{2}$
a) Elastic potential energy $=\frac{1}{2}K\frac{{A}^{2}}{4}=\frac{1}{4}\left(\frac{1}{2}K{A}^{2}\right)$
Kinetic Energy $=\frac{3}{4}\left(\frac{1}{2}K{A}^{2}\right)\to tal-P.E.$
Potential energy $=\frac{1}{4}x$ Total energy
Kinetic Energy $=\frac{3}{4}x$ total Energy
$KE=PE=\frac{1}{2}\left(\frac{1}{2}K{A}^{2}\right)=\frac{1}{2}K{\left(\frac{A}{\sqrt{2}}\right)}^{2}$
at position $x=\frac{A}{\sqrt{2}}$