potential energy

Question

A particle of mass 1.1 kg moves under the influence of a potential

U(x) = a/x + bx (a=3 and b=2.5). The particle's motion is restricted to the region x>0.

the force acting on the particle is a/x^2 - b

the point of equilibrium of the particle is 1.10 m

a) What is the minimal work required to move the particle from the equilibrium point to 5.48 m?

b) The particle is released from the rest at 5.48 m. What is the minimal distance from x=0 that it will reach? and what is its maximum velocity?

Answer 1

a) W = ΔU = U(5.48)-U(1.1) substitute the x values into the formula for U

b) The maximum velocity (if we assume that the equilibrium point is correct - the correct value is 1.0954...) is such that the change in potential energy (=W_ideal) = change in kinetic energy or

W(part a) = (1/2)m v_max² v_max = sqrt(2W/m)

What is the x position that it reaches on the other side of the equilibrium point?

That will be when the ΔU = W(a) = U(x)-U(1.1) and solve for quadratic in x.

I hope that helps.