Question

A 0.20-kg particle moves along the x axis under the in°uence of a stationary object. 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
B. 2:5 m=s
C. 5:7 m=s
D. 7:9 m=s
E. 11 m=s

ans: E

Not sure how to do this.

Answer 1

You sure this isn't kinetic energy? Potential energy is in terms of other variables, not velocity.

Answer 2

@raffle_snaffle I figured it out. You basically just use KE + PE = constant to solve.

Answer 3

Cool

Answer 4

Generally for a conservative energy field function \(U\), so you are assuming that it is a conservative field in this case, there is always a vector \(\vec F\) such that, \(\vec F = - \nabla U\), and for motion \(\vec F = m \vec {\ddot x }\)

in a 1-D system that simplifies to:

\(F= m \ddot x = - \dfrac{dU}{dx}\)

Write it as \(m \dot x \dfrac{d \dot x}{dx} = - \dfrac{dU}{dx}\)

Integrate:

\(\frac{1}{2} m \dot x^2 = - U + C\)

Apply the IV:

\(\frac{1}{2} (0.2) 5^2 = - 10 + C \implies C = 12.5 \)

\(\frac{1}{2} m \dot x^2 = - U(x) + 12.5\)

\( \dot x = \sqrt{ \dfrac{ 25 - 2U(x)}{ m} }\)

\( \dot x(0) = \sqrt{ \dfrac{ 25 - 0}{ 0.2} } = 11.18\) m/s

Answer 5

And that was completely OTT as i realised when i pushed the send button :(

once you make the assumption it's conservative, you write and solve \(\frac{1}{2} m \dot x^2 + U = C\) where C is the constant of the motion, in this case total energy

you then own it, as you did