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

This problem could be solved with or without using the calculus. Because the force is given, we assume that it must be solved without use of calculus. Otherwise, we can find the expression for the force by taking derivative with opposite sign from potential energy with respect to the coordinate. And find x for equilibrium position using properties of extremum of the function. BTW, in our case the equilibrium position corresponds to the minimum of potential energy of the body.

Let’s put the values of constants a and b into the equation for the potential energy. Then we have

                                                     U(x) = 3/x + 2.5x       

                   

a)    The minimum work required to move the particle from equilibrium position to point x = 5.48 is

                                       W = - Δ U

Or    W = - ((3/5.48 + 2.5 5.48) – (3/1.1 + 2.5 1.1) = - 8.77 J

We are assuming that a and b are measured in SI units.

b)   When particle released from rest at 5.48 m its total energy is the sum of kinetic and potential energies.

                                          E = K + U

The particle released from rest, hence K = 0 and total energy E = U(5.48).

Plugging in this value to formula for potential energy, we will get

                   ·      E = U = 3/5.48 + 2.5· 5.48 = 14.25 J               

In general

                            14.25 = 3/x + 2.5 x           

Or

                                           14.25 x = 3 + 2.5·x²

Putting this quadratic equation in standard form we get

                                      2.5 x² – 14.25 x + 3 = 0

It has 2 solutions x = 0.25 and x = 5.48

So, we see that the minimum distance from x = 0 is 0.25m

c)    The maximum velocity the particle will have in the equilibrium position  x = 1.10 m.

Applying the law of conservation of energy, we can write:

                                      E = K + U(1.10)

                                       14.25 = K + 3/1.1 + 2.5 ·1.1                   

And the maximum kinetic energy

                                                    K = 8.77 J

From the formula for kinetic energy

                                                          

                                                       K = mv²/2

maximum velocity

                                                     v = √2K/m

After substitution of values, we obtain

               

                                        v = 3.99 m/s

Please note that we did not use the equation for the force to do the problem.