# The force applied against a moving object travelling on a linear path is given by F(x)= sinx + 1 . How much work would it take to move the object over x in [ 0,pi/3] ?

Jan 13, 2018

3.08 units

#### Explanation:

Work done W= $F . \mathrm{ds}$ = $m \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{ds}$
= $m \cdot \mathrm{dv} \frac{\mathrm{ds}}{\mathrm{dt}}$ = $m {\left(\mathrm{dv}\right)}^{2}$
Given equation,
F = $\sin x + 1$
Or, $m v \frac{\mathrm{dv}}{\mathrm{dx}} = \sin x + 1$
Or, $m v \mathrm{dv} = \sin x \mathrm{dx} + \mathrm{dx}$
Integrating we get,
$m \left({\left(v 1\right)}^{2} - {\left(v 2\right)}^{2}\right) = \frac{- \cos x + x}{2}$
Putting the interval we get,
$m {\mathrm{dv}}^{2} = \left(- \left(\frac{1}{2}\right) + 1 + \left(\frac{\pi}{3}\right)\right) \cdot 2$ i.e 3.08 units

Jan 13, 2018

The work is $= 1.55 J$

#### Explanation:

The equation is

$\text{Work"="Force"xx"distance}$

Here,

The force is $F \left(x\right) = \sin x + 1$

Therefore,

$\Delta W = F \times \Delta x$

$\mathrm{dW} = \left(\sin x + 1\right) \mathrm{dx}$

Integrating both sides

$W = {\int}_{0}^{\frac{\pi}{3}} \left(\sin x + 1\right) \mathrm{dx}$

$= {\left[- \cos x + x\right]}_{0}^{\frac{\pi}{3}}$

$= \left(- \cos \left(\frac{\pi}{3}\right) + \frac{\pi}{3}\right) - \left(- \cos \left(0\right) + 0\right)$

$= - \frac{1}{2} + \frac{\pi}{3} + 1$

$= \frac{1}{2} + \frac{\pi}{3}$

$= 1.55 J$