Question

Variation of a force in a certain region is given by F = 6x² – 4x – 8. It displaces an object from x = 1 m to x = 2 m in this region. Calculate the amount of work done.

Answer 1

W = $$\underset{x=1}{\overset{x=2}{\int }}(6x^2-4x-8)$$

∴ W = $$\underset{x=1}{\overset{x=2}{\int }}6x^{2}dx-\underset{x=1}{\overset{x=2}{\int }}4x^{2}dx-\underset{x=1}{\overset{x=2}{\int }}8dx$$

${\displaystyle ={\left[\frac{6{\text{x}}^3}{3}\right]}_1^2-{\left[\frac{4{\text{x}}^2}{2}\right]}_1^2-{\left[\text{8x}\right]}_1^2}$

= (16 − 2) − (8 − 2) − (16 − 8) = 0

The work done is zero.