Respuesta :
Answer:
[tex]PE = \dfrac{0.45 r^4}{4} [/tex]
Explanation:
given,
proportionality constant = 0.45
F∝ r³
F = k r³
using conservation of energy change in potential energy is equal to work done.
d (PE) = F. dr
d (PE) = k r³. dr
integrating both side
[tex]\int {d (PE)} = k \int {r^3} dr[/tex]
[tex]PE = \dfrac{k r^4}{4} + C[/tex]
given that at r = 0; PE = 0
[tex]0 = \dfrac{k 0^4}{4} + C[/tex]
C = 0
[tex]PE = \dfrac{k r^4}{4} [/tex]
[tex]PE = \dfrac{0.45 r^4}{4} [/tex]
The magnitude of the force on a particle can be written as 0.45r³ when the particle is at a distance of r from the force center is [tex]PE = 0.45\dfrac{r^3}{4}[/tex].
What is the potential energy?
Potential energy is the energy that is within the object due to its position.
We know that the magnitude of the force on a particle increases as the cube of the distance from that point, that is, F∝ r3. therefore,
[tex]F \propto r^3\\\\F = kr^3[/tex]
We know that using conservation of energy change in potential energy is equal to work done.
[tex]d (PE) = F\cdot dr\\\\d (PE) = k r^3\cdot dr\\\\\text{Integrating both side}\\\\\int d (PE) = k\int r^3\cdot dr\\\\PE = k\dfrac{r^3}{4}+C\\\\[/tex]
As it is given that the value of the constant k is 0.45, therefore,
[tex]PE = k\dfrac{r^3}{4}+C\\\\PE = 0.45\dfrac{r^3}{4}+C[/tex]
As it is given that at radius r=0, the potential energy is 0, therefore,
[tex]0 = 0.45\dfrac{0^3}{4}+C\\\\C = 0[/tex]
Substitute the value of the Constant C as 0, therefore,
[tex]PE = 0.45\dfrac{r^3}{4}+0\\\\PE = 0.45\dfrac{r^3}{4}[/tex]
Hence, the magnitude of the force on a particle can be written as 0.45r³, when the particle is at a distance of r from the force center is [tex]PE = 0.45\dfrac{r^3}{4}[/tex].
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