Respuesta :
Answer
[tex]\Delta U= \alpha \frac{x^3}{3} \\[/tex]
Explanation:
given
[tex]F = -\alpha x^2 i[/tex]
where [tex]\alpha = 12 N/m^2[/tex]
now we know
[tex]\int\limits^W_0 {} \, dW = \int\limits^a_b {F.} \, dxi[/tex] ..................(i)
where dx is infinitesimal distance
[tex]W = \int\limits^a_b {-\alpha x^2} \, dx \\[/tex]
for x = a and b = 0
after integration we get
[tex]W = -\alpha \frac{x^3}{3}[/tex]
we know work done by conservative force will be equals to negative of potential energy
[tex]W = -\Delta U[/tex]
so we get
[tex]-\Delta U= -\alpha \frac{x^3}{3} \\\\\Delta U= \alpha \frac{x^3}{3} \\[/tex]
The potential energy function of the proton in terms of α and x is given by the equation ΔU = αx³/3.
Potential energy function :
Given a proton which is exerted upon by a force of F = -αx²î
where α = 12 N/m²
The change in potential energy ΔU of a system is given by the negative of the work done W by the system and vice-versa.
Now, Work done:
[tex]W = -\int\limits^{}_{} {F(x)} \, dx\\\\W = -\int\limits^{}_{} {\alpha x^2} \, dx\\\\W=-\alpha \frac{x^3}{3}+C[/tex]
ΔU = -W
ΔU = [tex]\alpha \frac{x^3}{3}-C[/tex]
given that U = 0 at x = 0
⇒ C = 0
So;
ΔU = αx³/3 is the required function.
Learn more about potential energy:
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