Calculating F, V and C: Why Can It Not Be Obtained This Way?

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AI Thread Summary
The discussion centers on the incorrect approach to calculating force, velocity, and potential energy in a specific problem. The error lies in treating acceleration as the derivative of velocity with respect to position instead of time, leading to an incorrect expression for velocity. The correct method involves recognizing that acceleration is the time derivative of velocity and requires solving a differential equation using separation of variables. Additionally, a different notation for derivatives can clarify the relationships between acceleration, velocity, and position. The conversation emphasizes the importance of correctly applying calculus principles to derive accurate physical equations.
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Homework Statement
A 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by

U(x) = (8.0 J/m^2)x^2 + (2.0 J/m^4)x^4,

where x is in coordinate of the particle. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is:

a) 0 m/s
b) 2.5 m/s
c) 5.7 m/s
d) 7.9 m/s
e) 11 m/s

The answer is E.
Relevant Equations
K1 + U1 = K2 + U2
The correct answer can be obtained by the calculation as attached.
12.png

However it can not be gotten by the following way. Why?
F = -∇U = -[ 16x + 8x^3] = ma
Since m = 0.2, a = -80x - 40x^3
V = -40x^2 - 10x^4 +C =5
c= 50 + 5 =55
 
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You have treated acceleration as if it were the derivative of velocity with respect to position when it is the derivative of velocity with respect to time.
 
Your mistake is that the velocity is the integral of acceleration with respect to time t , not with respect to the distance or x-coordinate. So your line

hidemi said:
V=-40x^2-10x^4+C
is wrong. To continue with your way, you should set ##a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v## and solve the ODE $$-16x-8x^3=m\frac{dv}{dx}v$$ by the separating variables technique.
 
Delta2 said:
Your mistake is that the velocity is the integral of acceleration with respect to time t , not with respect to the distance or x-coordinate. So your lineis wrong. To continue with your way, you should set ##a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v## and solve the ODE $$-16x-8x^3=m\frac{dv}{dx}v$$ by the separating variables technique.
Thanks for your remider :)
 
Similar technique, but slightly different on execution is to let a dot denote derivative with respect to time, i.e., ##a = \ddot x## and ##v = \dot x##. The equation of motion becomes
$$
\ddot x = -80x - 40 x^3.
$$
Multiplying both sides with ##v = \dot x## leads to
$$
v \dot v = \ddot x \dot x = -80 x \dot x - 40 x^3 \dot x.
$$
Noting that for any function ##g(t)##, it holds that ##d(g^n)/dt = n g^{n-1} \dot g## therefore leads to
$$
\frac{d}{dt} \left[\frac 12 v^2\right] = \frac{d}{dt}\left[ - 40 x^2 - 10 x^4\right],
$$
which means the expressions being differentiated differ by a constant, which is essentially the conservation of energy equation.
 
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