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Question 1 1 pts Given the following Potential Energy function with depends on the position , PE (k/2) x3 what is the force in the x direction? O F (3 k/2) x2 O F-(k/2) x2 O F(3 k/2) x2

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Answer #1

Potential energy U = (k/2) x^3

F = -dU/dx

F = - d((K/2)x^3) / dx

= (-3k/2) x^2

so F = -(3k/2)x^2

so last option is the correct answer

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