Respuesta :
Answer:
[tex]F_{x} (-0.80) = 1.29 N[/tex]
The force acting on the particle at x = -0.800 has magnitude of 1.29 N and directed toward the origin
Explanation:
First we write the potential energy produced by the force
[tex]U(x) = ax^{4}i[/tex]
where α = 0.630 [tex]N/m^{4}[/tex]
therefore [tex]U(x)=0.630x^{4}i[/tex]
For motion along a straight line, a conservative force is the negative derivative of its associated potential energy function U(x):
[tex]F_{x}(x) = -\frac{d}{dx} (0.630x^{4}i )[/tex]
[tex]F_{x}(x) = -0.630\frac{d}{dx} (x^{4} )i = -0.630 (4x^{3} )i[/tex]
therefore [tex]F_{x} (x) = -2.52x^{3} i[/tex]
At x = -0.80 the force equals
[tex]F_{x}(-0.630) = - 2.53 X (-0.80)^{3}i } = 1.29 N[/tex]
[tex]F_{x} (-0.80) = 1.29 N[/tex]
The magnitude of the force is 1.29 N in the direction directed towards the origin.
From the given information, the relationship between the force to the potential energy for the conservative system can be computed as:
[tex]\mathbf{F = -\dfrac{dU(x)}{dx} }[/tex]
The function of the force producing the potential energy can be expressed as:
[tex]\mathbf{U(x) = ax^4}[/tex]
where;
- the constant ∝ = 0.630 J/m⁴
The force that pulls on a particle that moves in the parallel direction to the x-axis is as follows;
[tex]\mathbf{F = -\dfrac{dU(x)}{dx} }[/tex]
[tex]\mathbf{F = -\dfrac{d(ax^4)}{dx} }[/tex]
[tex]\mathbf{F = -\alpha (\dfrac{d}{dx}x^4) }[/tex]
[tex]\mathbf{F = -\alpha (4x^3) }[/tex]
∴
The magnitude of the force acting on the particle is located at x = 0.800 m is:
[tex]\mathbf{\Big|F\Big|_{x=0.800 \ m} =\Big |-\alpha (4x^3)\Big |}[/tex]
[tex]\mathbf{\Big|F\Big|_{x=0.800 \ m} =\Big |-(0.630 \ J/m^4) (4(0.800 \ m)^3)\Big |}[/tex]
[tex]\mathbf{\Big|F\Big|_{x=0.800 \ m} =1.29 \ N}[/tex]
Therefore, we can conclude that magnitude of the force is 1.29 N in the direction directed towards the origin.
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