This problem involves analyzing the relationship between the distance traveled by two objects subject to forces of different magnitudes. The force acting on both objects is position-dependent, described by the equation:

$\vec{F} = (Ax - B)\hat{i}$

Where $A$ and $B$ are constants, $x$ is the position, and $\hat{i}$ is the unit vector in the direction of motion.

Key points:

• Initially, the object starts at rest.
• The first object experiences a force $F$ and moves a distance $D_0$ before coming to rest.
• The second identical object experiences a force $2F$ and comes to rest after moving a distance $D_1$.

Step 1: Work-Energy Principle

We will apply the work-energy theorem, which states:

$W = \Delta K$

The change in kinetic energy ($\Delta K$) is zero because both objects start and end at rest. Thus, the total work done by the force on the object is equal to the initial potential energy. The work done by a variable force can be written as:

$W = \int_{0}^{D} F(x) \, dx$

Substitute the force $F(x) = (Ax - B)$:

$W_0 = \int_{0}^{D_0} (Ax - B) \, dx$

$W_0 = \left[\frac{Ax^2}{2} - Bx \right]_{0}^{D_0} = \frac{A D_0^2}{2} - B D_0$

For the second object, which experiences twice the force ($2F$):

$W_1 = \int_{0}^{D_1} 2(Ax - B) \, dx$

$W_1 = 2 \left[\frac{Ax^2}{2} - Bx \right]_{0}^{D_1} = A D_1^2 - 2 B D_1$

Step 2: Relationship Between $D_0$ and $D_1$

Since the work-energy principle applies to both objects, and both objects have the same initial and final kinetic energy (zero), the work done on both objects is proportional to the force applied.

Thus, the ratio of the distances can be found by equating the total work done for each case:

$\frac{A D_0^2}{2} - B D_0 = A D_1^2 - 2 B D_1$

This equation will give the relationship between $D_0$ and $D_1$. Solving this equation would show the exact ratio, but we can expect $D_1 > D_0$, since the second object experiences a larger force and hence would move a greater distance before coming to rest.

Would you like to explore the solution in more depth or solve the equation explicitly? Here are five additional questions that could help expand on this concept:

1. What is the significance of the force being a linear function of position?
2. How does the work-energy principle apply when forces vary with position?
3. What would happen if the force increased exponentially instead of linearly with position?
4. How can the relationship between force and distance change if the initial velocities were not zero?
5. How does the mass of the object affect the total work done in this scenario?

Tip: When dealing with varying forces, always integrate the force over the distance to calculate the work done, since simple multiplication won't suffice.