Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007)

Chapter 2: Integration; Section 2: Definite Integrals; page 2

Applications, Exercise 3

Question

For each of the following forces F which depend on the x coordinate, and are applied to a mass point m, use the given initial conditions in order to

Reminder

... the second law of Newton for the motion of a mass point m in one dimension with a force F dependent on the coordinate x only is

where v is the velocity and t - the time.

... we'll call  U  the potential energy.

One can obtain the total energy from the initial conditions ... as

which represents a very useful relation between v² and the coordinate x.

From the energy conservation and the direction of the force alone, one can obtain a qualitative description of the motion. One has to keep in mind that at the point of vanishing velocity, the motion will continue in the direction of the force (which is also the direction of the acceleration).

Parts 1-3

Solution of question 1

1. The potential energy for F=−ax³ is
   
   $$U\left(x\right)=a\int x^3\text{d}x=\frac{a{x}^4}{4}$$


2. For the given initial conditions the expression of energy conservation is
   
   $$\frac{m{v}^2}{2}+\frac{a{x}^4}{4}=\frac{a{x}_0^4}{4}$$


3. Qualitative description of the motion
• The motion starts at x=x_o from rest.
• Continues accelerating in the negative direction of x.
• At x=0 the velocity is $v=-x_0^2\sqrt{\frac{a}{2m}}$.
• After x=0 the velocity decelerates because the force switches sign.
• At x=−x_o the mass is at rest
• From x=−x_o the motion continues with accelerating velocity in the positive direction of x.
• At x=0 the velocity is maximal: $v=x_0^2\sqrt{\frac{a}{2m}}$.
• After x=0 the mass continues to move in the positive x direction, but decelerating because of the change in sign of the force.
• At x=x_o the mass is at rest.
• From x=x_o the motion continues as described at the beginning.

Parts 4-6

Solution of question 2

4. The potential energy for F=kx is
   
   $$U\left(x\right)=-k\int x\text{d}x=-\frac{k{x}^2}{2}$$


5. For the given initial conditions the expression of energy conservation is
   
   $$\frac{m{v}^2}{2}-\frac{k{x}^2}{2}=-\frac{k{x}_0^2}{2}$$


6. Qualitative description of the motion
• The motion starts at x=x_o from rest.
• From rest the mass accelerates in the positive direction of x in a way that the increase of v² is compensated by by the growing x² according to part 5.
• The mass will continue its accelerated motion till infinity.

Parts 7-9

Solution of question 3

7. The potential energy for F=bx⁻² is
   
   $$U\left(x\right)=-b\int \frac{\text{d}x}{x^2}=\frac{b}{x}$$


8. For the given initial conditions x_o>0 and v_o<0 the expression of energy conservation is
   
   $$\frac{m{v}^2}{2}+\frac{b}{x}=\frac{m{v}_0^2}{2}+\frac{b}{x_0}$$


9. Qualitative description of the motion
• The motion starts at x=x_o with velocity v_o in the negative x direction.
• The mass descelerates since the force is positive.
• The mass is at rest (v=0) at
  
  $$x=x_0\frac{2b}{m{x}_0{v}_0^2+2b}$$
  which is positive and is obtained from part 8.
• From rest the mass accelerates in the positive direction of x.
• At x=x_o the velocity becomes v=|v_o| (positive).
• From that point the mass continues its acceleration till infinity.

Score

If the final score is not an integer, subtract half point.

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