an object that is 0.5 m above the ground has the same amount of potential energy as a spring that is…
an object that is 0.5 m above the ground has the same amount of potential energy as a spring that is stretched 0.5 m. each distance is then doubled. how will the potential energies of the object and the spring compare after the distances are doubled? the gravitational potential energy of the object will be two times greater than the elastic potential energy of the spring. the elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object. the elastic potential energy of the spring will be two times greater than the gravitational potential energy of the object. the potential energies will remain equal to one another.
Answer
Explanation:
Step1: Recall potential - energy formulas
The gravitational potential energy formula is $U_g = mgh$, and the elastic potential energy formula is $U_s=\frac{1}{2}kx^2$. Initially, $mgh_1=\frac{1}{2}kx_1^2$ with $h_1 = 0.5m$ and $x_1 = 0.5m$.
Step2: Calculate new potential energies
When the distances are doubled, $h_2 = 1m$ and $x_2 = 1m$. The new gravitational potential energy $U_{g2}=mgh_2 = mg\times1$. The new elastic potential energy $U_{s2}=\frac{1}{2}kx_2^2=\frac{1}{2}k\times1^2$. Since $mgh_1=\frac{1}{2}kx_1^2$ and $h_2 = 2h_1$, $x_2 = 2x_1$, we have $U_{g2}=2mgh_1$ and $U_{s2}=4\times\frac{1}{2}kx_1^2$. Because $mgh_1=\frac{1}{2}kx_1^2$, $U_{s2}=4\times mgh_1$ and $U_{g2}=2mgh_1$. So $U_{s2} = 2U_{g2}$.