A particle moving on the x-axis experiences a force given by Fx = qx², where q is a constant. How much work is done on the particle as it moves from x=0 to x=d?

Verified step by step guidance

1. Identify the force function and the limits of integration. In this case, the force function is given as Fx = qx^2, and the particle moves from x = 0 to x = d.
2. Recall the work done by a force in moving an object along the x-axis is given by the integral of the force function with respect to x over the limits of motion. The formula for work done, W, is W = \int_{x_1}^{x_2} F(x) dx.
3. Substitute the given force function into the work formula. This gives W = \int_{0}^{d} qx^2 dx.
4. Perform the integration of the function qx^2 from 0 to d. The integral of x^2 is \frac{1}{3}x^3. Therefore, the integral becomes \frac{q}{3}x^3 evaluated from 0 to d.
5. Evaluate the definite integral by substituting the upper and lower limits of integration. This results in W = \frac{q}{3}d^3 - \frac{q}{3}(0)^3 = \frac{q}{3}d^3.

Key Concepts

Work Done by a Force

Work is defined as the integral of force over the distance moved in the direction of the force. Mathematically, it is expressed as W = ∫ F dx. In this case, the force Fx = qx² varies with position, so the work done as the particle moves from x=0 to x=d requires evaluating the integral of this force function over that interval.

Integration in Physics

Integration is a fundamental mathematical tool used in physics to calculate quantities that accumulate over a continuous range. In the context of work, it allows us to sum up the infinitesimal contributions of force over the distance traveled. For the given force function, integrating qx² from 0 to d will yield the total work done on the particle.

Force as a Function of Position

In this scenario, the force acting on the particle is not constant but varies with position, specifically as Fx = qx². This means that the force increases with the square of the position x. Understanding how force varies with position is crucial for accurately calculating work, as it directly influences the integral that determines the total work done.