Question

How much work is required to move an object from \(x=0\) to \(x=3\) (measured in meters) in the presence of a force (in \(\mathrm{N}\) ) given by \(F(x)=2 x\) acting along the \(x\) -axis?

Short answer

Answer: The work done on the object is 9 Joules.

Step-by-step solution

Write down the formula for work done

To calculate the work done when moving the object, we use the formula: Work = \(\int_{x_1}^{x_2} F(x) dx\) where \(F(x)\) is the force function and x1 and x2 are the initial and final positions of the object on the x-axis.

Identify the given Force function and the limits of integration

We are given the force function as: \(F(x) = 2x\) We are also given the initial and final positions of the object: Initial position (\(x_1\)): 0m Final position (\(x_2\)): 3m

Set up the integral for work done

Using the given information, we can set up the integral for work done: Work = \(\int_{0}^{3} 2x dx\)

Evaluate the integral

Now, evaluate the integral: Work = \(2\int_{0}^{3} x dx=\left[x^2\right]_0^3\)

Calculate work done

Plug in the limits of integration to get the numerical value for work done: Work = \((3^2) - (0^2) = 9 - 0 = 9\)

Interpret the result

The work required to move the object from x=0 to x=3 in the presence of the given force is 9 Joules.