The force applied against a moving object travelling on a linear path is given by F(x)=3x^2+x . How much work would it take to move the object over x in [1,3 ] ?

Mar 3, 2018

here,work will be done against the force to move the object,

so,work done $\mathrm{dW} = F \mathrm{dx}$ (as,angle between $F$ and $\mathrm{dx}$ is zero,as moving along a linear pathway)

so, $\mathrm{dW} = \left(3 {x}^{2} + x\right) \mathrm{dx}$

or, $\mathrm{dW} = 3 {x}^{2} \mathrm{dx} + x \mathrm{dx}$

so, ${\int}_{0}^{W} = 3 {\int}_{1}^{3} {x}^{2} \mathrm{dx} + {\int}_{1}^{3} x \mathrm{dx}$

so, $W = {\left[{x}^{3}\right]}_{1}^{3} + \frac{1}{2} {\left[{x}^{2}\right]}_{1}^{3} = 30 J$

Mar 3, 2018

$W = 30$ J

Explanation:

Simply integrate using the bounds provided:
$W = \setminus {\int}_{1}^{3} F \left(x\right) \mathrm{dx}$

$= \setminus {\int}_{1}^{3} \left(3 {x}^{2} + x\right) \mathrm{dx}$

$= 3 \setminus {\int}_{1}^{3} {x}^{2} \mathrm{dx} + \setminus {\int}_{1}^{3} x \mathrm{dx}$

$= {\left({x}^{3} + \frac{1}{2} {x}^{2}\right)}_{x = 1}^{x = 3}$

$= 27 + \frac{9}{2} - \left(1 + \frac{1}{2}\right)$

$= 30$ J