To tackle the problem of a particle moving along the +x axis under the influence of a force derived from a given potential, we can break it down into three parts: finding the expression for the force, determining the interval for the particle when its total energy is zero, and calculating the maximum kinetic energy of the particle. Let’s go through each part step by step.
Finding the Expression for Force F(x)
The force acting on a particle in a potential field is given by the negative gradient of the potential energy. Mathematically, this is expressed as:
F(x) = -dU/dx
Given the potential energy function:
U(x) = ax³ + bx
We can differentiate this with respect to x:
- First, differentiate ax³: d(ax³)/dx = 3ax²
- Next, differentiate bx: d(bx)/dx = b
Putting it all together, we have:
F(x) = - (3ax² + b)
Determining the Interval for the Particle
Next, we need to find the interval where the particle can be trapped when its total energy is zero. The total mechanical energy (E) of the particle is the sum of its kinetic energy (K) and potential energy (U):
E = K + U
When the total energy is zero, we have:
0 = K + U
This implies:
K = -U
Since kinetic energy cannot be negative, we need U to be non-positive:
U(x) = ax³ + bx ≤ 0
To find the values of x1, we need to solve the inequality:
ax³ + bx ≤ 0
Factoring out x, we get:
x(ax² + b) ≤ 0
This inequality holds true when:
- Both factors are zero, or
- One factor is positive and the other is negative.
Setting the expression equal to zero gives us:
x(ax² + b) = 0
This results in:
- x = 0
- ax² + b = 0 → x² = -b/a (valid if a and b have opposite signs)
Thus, the value of x1 can be expressed as:
x1 = √(-b/a) (assuming a < 0 and b > 0 for the particle to be trapped).
Calculating Maximum Kinetic Energy
To find the maximum kinetic energy of the trapped particle, we can use the relationship we established earlier:
K = -U
At the boundary where the particle is trapped (x = x1), we substitute x1 into the potential energy function:
U(x1) = a(x1)³ + b(x1)
Substituting x1 = √(-b/a) into U(x):
U(√(-b/a)) = a(√(-b/a))³ + b(√(-b/a))
Calculating this gives:
U(√(-b/a)) = a(-b/a)√(-b/a) + b√(-b/a)
Which simplifies to:
U(√(-b/a)) = -b√(-b/a) + b√(-b/a) = 0
Thus, the maximum kinetic energy is:
K_max = -U(x1) = -0 = 0
However, to find the maximum kinetic energy at the point just before reaching x1, we can evaluate U at a point slightly less than x1, leading to:
K_max = -U(x) = -[a(x)³ + b(x)]
In summary, the maximum kinetic energy of the trapped particle can be expressed in terms of a and b as:
K_max = -U(x1) = -[a(√(-b/a))³ + b(√(-b/a))]
In conclusion, we have derived the force expression, determined the trapping interval, and calculated the maximum kinetic energy of the particle. Each step builds on the principles of classical mechanics and the relationships between force, potential energy, and kinetic energy.
