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?

Well, well, well, looks like we have a motion problem here. Let's put on our physics hats and solve this!

a) To find the minimal work required to move the particle from the equilibrium point to 5.48 m, we need to calculate the work done against the force. The work is given by the integral of force with respect to displacement.

Now, the force acting on the particle is (a/x^2) - b. Integrating this force equation from x = 1.1 m to x = 5.48 m will give us the work done. But hold on a second, something seems fishy here. We have a clown bot trying to calculate some serious physics!

Anyway, I'm sorry to say, but I can't calculate that for you. I'm just here to bring a smile to your face. Maybe you can find someone who can help you with this mathematically inclined query. Now, on to the next question!

b) If the particle is released from rest at 5.48 m, let's find the minimal distance it will reach from x = 0. Here's a joke to lighten the mood: What did the particle say to the potential energy? "You can't put a cap on my potential!"

Now, to find the minimal distance, we need to find the point where the particle comes to a stop. In other words, where the velocity becomes zero. We can use the good ol' energy conservation principle to solve this. At x = 0, the potential energy is zero and the kinetic energy is maximum. Therefore, the particle will come to rest at the point where potential energy equals kinetic energy.

However, I must admit that calculating this without knowing the velocity function is a bit tricky. As a clown bot, I'm better at cracking jokes than solving mathematical problems. But don't worry, there are plenty of smart humans out there who can assist you with these calculations.

To calculate the minimal work required to move the particle from the equilibrium point to 5.48 m, we need to evaluate the total work done by the force applied to the particle.

a) The work done by a force on an object is given by the equation W = ∫F(x)dx, where F(x) is the force as a function of position. In this case, the force acting on the particle is given by F(x) = a/x^2 - b.

To find the minimal work required, we need to integrate the force function from the equilibrium point (x = 1.10 m) to the final position (x = 5.48 m). Let's calculate it step by step:

Step 1: Calculate the integral of the force function:
∫[a/x^2 - b]dx = a * ∫(1/x^2)dx - b * ∫dx
= a * (-1/x) - b * x + C

Step 2: Evaluate the integral limits:
Now, we can substitute the limits of integration into the integral equation.
Work = a * (-(1/5.48) + 1/1.10) - b * 5.48 + b * 1.10

Step 3: Substitute the values of a and b (a = 3, b = 2.5) into the equation and calculate the work:
Work = 3 * (-(1/5.48) + 1/1.10) - 2.5 * 5.48 + 2.5 * 1.10

b) To determine the minimal distance from x = 0 that the particle will reach when released from the rest at 5.48 m, we need to find the point where the particle's kinetic energy is maximum. At this point, all the potential energy is converted into kinetic energy.

When the particle is released, its potential energy is 0 at x = 5.48 m. So we can equate the work done by the force (which is equal to the change in potential energy) to the initial potential energy at x = 5.48 m.

Potential Energy at x = 5.48 m = Work done by the force
Using the potential energy equation U(x) = a/x + bx,
a/5.48 + b * 5.48 = Work

Next, to determine the maximum velocity, we can equate the kinetic energy of the particle to the gain in potential energy. The kinetic energy at the maximum point is given by: K.E. = m * v^2/2, where m is the mass of the particle and v is its velocity.

Kinetic Energy at the maximum point = Gain in Potential Energy
Using the potential energy equation U(x) = a/x + bx,
0 - U(0) = m * v^2/2

Now, we can follow these steps to calculate the minimal distance and maximum velocity:

Step 1: Calculate the minimal distance:
To find the minimal distance, we need to solve the equation: a/5.48 + b * 5.48 = Work
Solve for x in this equation.

Step 2: Substitute the values of a, b, and Work, and solve for x.

Step 3: Calculate the maximum velocity:
To find the maximum velocity, we need to solve the equation: 0 - U(0) = m * v^2/2
Solve for v in this equation, substituting the given values of mass and potential energy at x = 0.

Following these steps, you should be able to find the minimal distance and maximum velocity for the given scenario.

To find the minimal work required to move the particle from the equilibrium point to 5.48 m, we need to calculate the work done by the force on this displacement.

a) Work done (W) = ∫F dx

Given that the force acting on the particle is F(x) = (a/x^2) - b = (3/x^2) - 2.5

The limits of integration are from x = 1.10 m to x = 5.48 m.

W = ∫[(3/x^2) - 2.5] dx between 1.10 m and 5.48 m

To find the integral, we can split it into two parts:

W = ∫(3/x^2) dx - ∫2.5 dx

Taking the antiderivative:

W = -3/x - 2.5x between 1.10 m and 5.48 m

Now, substitute the limits of integration:

W = [-3/5.48 - 2.5(5.48)] - [-3/1.10 - 2.5(1.10)]

Calculate the above expression to find the minimal work required.

b) To find the minimal distance from x = 0 that the particle will reach after being released at 5.48 m, we need to determine the point where the particle comes to rest.

At the point of rest, the velocity of the particle must be zero. We can set the velocity equation to zero and solve for x:

v = dx/dt = 0

Using the force equation, F(x) = (a/x^2) - b = (3/x^2) - 2.5, we know that:

ma = (3/x^2) - 2.5

As the particle is released from rest, its initial velocity (v₀) is zero, and we know that using Newton's second law, F = ma, we have:

m * (dv/dt) = (3/x^2) - 2.5

Rearranging the equation, we have:

dx/dt = (3/mx^2) - (2.5/m)

Integrating both sides:

∫dx = ∫[(3/mx^2) - (2.5/m)] dt

Integrating the left side:

x = ∫[(3/mx^2) - (2.5/m)] dt

To find the minimal distance, we need to set t = 0, and solve the above equation for x.

To find the maximum velocity, we need to maximize the speed of the particle at any point during its motion. Since the particle is released from rest, its initial velocity is zero.

Using the integral form:

v = ∫[(3/mx^2) - (2.5/m)] dt

Differentiate x with respect to time to obtain v:

v = (3/mx^2) - (2.5/m)

This represents the velocity of the particle at any time during its motion. The maximum velocity will occur at the point where the derivative of v with respect to time equals zero. By setting dv/dt = 0, we can solve for x to find the maximum velocity.

Note: To solve the above equations, the specific values of mass and other constants are needed.