Force of Particle at x=-0.660 m | Alpha = 1.50 J/m^4

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To find the force acting on the particle at x = -0.660 m, the relationship between force and potential energy is utilized, specifically that force is the negative derivative of potential energy with respect to position. Given the potential energy function U(x) = alpha x^4, where alpha = 1.50 J/m^4, the force can be calculated using F(x) = -dU/dx. This results in F(x) = -4 * alpha * x^3. Substituting the values, the force at x = -0.660 m can be determined. Understanding this relationship is crucial for solving similar problems in mechanics.
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Homework Statement


A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy U(x) given by U(x)= alpha x^4 where alpha=1.50 J/m^4.

What is the force when the particle is at position x = -0.660 m?


Homework Equations





The Attempt at a Solution


I know this has to do with derivaties, a the answer will comes from 4*1.5*some other number cubed. I just don't know what.
 
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