Use Newtonian dynamics to find the speed at this position

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The discussion focuses on solving a homework problem involving a non-Hookean spring with specific force coefficients. Participants are tasked with calculating the potential energy function U(x) using the integral of the force F, which is expressed as F = αx + βx². They also need to determine the speed of a mass attached to the spring when it is displaced from equilibrium and released, applying conservation of energy principles. Additionally, the instantaneous power at a specific position is to be calculated, with some uncertainty expressed about the application of Newtonian dynamics. The conversation emphasizes the integration of force and energy conservation methods to arrive at the required solutions.
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I need some help with this hwk! Please someone help me...:confused:
A certain spring is found not to obey Hook’s law; it exerts a restoring force if it is stretched or compressed, where α = 60.0 N / m and β = 18.0 N / m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for the spring. Let U = 0 when x = 0. (b) An object with mass m = 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the + x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.500 m to the right of the equilibrium position? (c) Use Newtonian dynamics to find the speed at this position. (d) What is the instantaneous power when x = 0.500 m?
 
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According to the given coefficients it seems the relationship is of the form

F = \alpha x + \beta x^2

therefore

(a) use

U(x) = \int {F\ dx} + C

(b) use conservation of energy

(c) I am not sure what to do here maybe

a = \frac{dv}{dt} = \frac{F}{m}

therefore

\frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx}v = \frac{F}{m}

giving

\frac{1}{2}v^2 = \int {\frac{F}{m}dx} + C

which in essense is conservation of energy again ?
 
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thankyou...
 
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