]> Exercise 3

# 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

• obtain the expression of energy conservation
• describe qualitatively the motion
1. F=−ax³ where a is a positive constant and the initial conditions are xo>0 and vo=0.
2. F=kx where k is a positive constant and the initial conditions are xo>0 and vo=0.
3. F=bx−2 where b is a positive constant and the initial conditions are xo>0 and vo<0.

## 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

 $m\frac{\text{d}v}{\text{d}t}=F\left(x\right)$ (2.2.2.14)

where v is the velocity and t - the time.

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

 $\int F\left(x\right)\text{d}x=-U\left(x\right)$ (2.2.2.16)

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

 $\frac{m{v}^{2}}{2}+U\left(x\right)=E$ (2.2.2.18)

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=xo 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=−xo the mass is at rest
• From x=−xo 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=xo the mass is at rest.
• From x=xo the motion continues as described at the beginning.

## Parts 4-6

Solution of question 2

1. The potential energy for F=kx is

$U\left(x\right)=-k\int x\text{d}x=-\frac{k{x}^{2}}{2}$

2. For the given initial conditions the expression of energy conservation is

$\frac{m{v}^{2}}{2}-\frac{k{x}^{2}}{2}=-\frac{k{x}_{0}^{2}}{2}$

3. Qualitative description of the motion
• The motion starts at x=xo 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

1. The potential energy for F=bx−2 is

$U\left(x\right)=-b\int \frac{\text{d}x}{{x}^{2}}=\frac{b}{x}$

2. For the given initial conditions xo>0 and vo<0 the expression of energy conservation is

$\frac{m{v}^{2}}{2}+\frac{b}{x}=\frac{m{v}_{0}^{2}}{2}+\frac{b}{{x}_{0}}$

3. Qualitative description of the motion
• The motion starts at x=xo with velocity vo 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=xo the velocity becomes v=|vo| (positive).
• From that point the mass continues its acceleration till infinity.

## Score

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

By parts:

Parts 1,4,7 are worth half point each.
Parts 2,5,8 are worth 1 points each.
Parts 3,6,9 are worth 2 points each.

By questions:

Questions 1,2,3 are worth 3 and half points each.