Use Newtonian dynamics to find the speed at this position

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SUMMARY

This discussion focuses on applying Newtonian dynamics to solve a problem involving a non-Hookean spring characterized by coefficients α = 60.0 N/m and β = 18.0 N/m². The potential-energy function U(x) is derived using the force equation F = αx + βx², leading to U(x) = ∫F dx. The speed of a 0.900 kg mass attached to the spring is calculated at a position 0.500 m from equilibrium using conservation of energy principles. Instantaneous power at this position is also explored, reinforcing the relationship between force, mass, and acceleration.

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  • Basic principles of conservation of energy
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Jessikalinphy
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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 ?
 
Last edited:
thankyou...
 

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