Q.A particle moves along x-axis from x = 0 to x = 5 m under the influence of force F = 5 - 4x + 3x^2. Find the work done in this process.

Since F = (2+x) N varies with x, the work done is found by integrating F dx from x=1 m to x=2 m, giving 3.5 J.

For a variable force, work done is:

$W = \int_{x_1}^{x_2} F\,dx$

Here $F = (2+x)\ N$, $x_1 = 1\ m$, $x_2 = 2\ m$:

$W = \int_{1}^{2}(2+x)\,dx = \left[2x + \dfrac{x^2}{2}\right]_1^2$

At $x=2$: $2(2) + \dfrac{(2)^2}{2} = 4 + 2 = 6$

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Work done by a position-dependent force is W = integral of F dx; evaluating the integral from x = 0 to x = 5 m gives W = 100 J.

Given F = 5 - 4x + 3x^2 (in newtons, x in metres).

Work done, W = integral from 0 to 5 of F dx = integral from 0 to 5 of (5 - 4x + 3x^2) dx

= [5x - 2x^2 + x^3] evaluated from 0 to 5

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