A force with magnitude \(F=a \sqrt{x}\) acts in the \(x\) -direction, where \(a=9.5 \mathrm{N} / \mathrm{m}^{1 / 2} .\) Calculate the work this force does as it acts on an object moving from (a) \(x=0\) to \(x=3.0 \mathrm{m} ;\) (b) \(3.0 \mathrm{m}\) to \(6.0 \mathrm{m}\) and (c) \(6.0 \mathrm{m}\) to \(9.0 \mathrm{m}\)

When it comes to physics, variable force integration is a crucial concept that helps us calculate the amount of work done by a force that changes with position. Unlike constant forces that apply the same amount of force throughout the motion, variable forces require an understanding of calculus to determine the total work done.

The concept is straightforward: you have a force that depends on the position, such as the provided square root force function, and you want to find the total work done by this force over a certain distance. To do this, we set up an integral where the force function is multiplied by an infinitesimal displacement, denoted as 'dx', and we sum up all these tiny amounts of work for the entire range of motion.

To convert your understanding into mathematical terms, the equation for work done by a variable force from position \(x_1\) to \(x_2\) is: \[W=\int_{x_1}^{x_2} F dx\]. Here, \(F\) is the variable force that changes with position. This integration process is what allows us to accumulate all the tiny bits of work over the range and provides us with the total work done by the force.

A square root force function such as \(F=a \sqrt{x}\) has a particular significance in physics problems. It is a type of variable force wherein the force applied varies with the square root of the distance or position. This means as the object moves, the applied force increases at a rate proportional to the square root of the distance it has traveled.

Mathematically, whenever you encounter a square root function during an integration, it implies the function has a half exponent, as in \(x^{1/2}\). When integrated, you apply the power rule which increases the exponent by one and then divide by the new exponent.

Integration Example

For the square root function \(F=a \sqrt{x}\), the integration proceeds as follows: \[W = a \int \sqrt{x} dx = a \cdot \frac{2}{3} x^{3/2}\]. It is also important to remember that such functions lead to forces that increase less rapidly than linear functions as the object moves further away from the starting point.

The term work in physics refers to the transfer of energy when a force is applied over a distance. The standard unit for measuring work is the Joule (J). The task of calculating work in physics often involves understanding the relationship between force and displacement.

For constant forces, the calculation is simple: it is the product of the magnitude of the force and the distance moved in the direction of the force. However, when dealing with variable forces, we need to rely on calculus, specifically integration, to find the work done over a range of positions. In the given exercise, we saw that the work done is a function of the distance through which the force acts, and since the force varies with the square root of distance, the integration yields differing amounts of work for different segments of the motion.

To put this in context with our exercise, as the object moves from one interval to another, we notice that the force's effect becomes greater, hence more work is being done, and with the proper integration, we can accurately calculate that work, irrespective of how the force changes with distance.