Question

Find the work done by the force field F(x, y) = xi + (y + 6)j in moving an object along an arch of the cycloid r(t) = (t − sin(t))i + (1 − cos(t))j, 0 ≤ t ≤ 2π.

Answer 1

177.65

Step-by-step explanation:

Work done by the force field,F along path C is given by:

`W=\int\limits_C F.dr`

Given that:

`r(t) = (t -sin(t)i + (1 -cos(t)j,\\\\(dr)/(dt)=(1-cos t)i+sin t\ j\\\\dr=(1-cos\ t)i +sin\ t \ j)dt\\\\F(x,y)=x\ i +(y+6)j\\\\F(r(t))=(t-sin \ t) i+((1-cos \t)+2)j\\\\\\`

`F(r(t))=(t-sin \ t)i+(3-cos \ t)j\\\\\\W=\int\limits_C F.dr\\=\int\limits^(6\pi)_0(t-sin \ t)i+(3-cos \ t)j).((1-cos \ t)i+sin \ t \ j)dt\\\\=\int\limits^(6\pi)_0(t-sin \ t)(1-cos \ t)+(3-cos \ t)sin \ t \ dt\\\\=\int\limits^(6\pi)_0t-tcos \ t+2sin\ t\ dt\\\\=\int\limits^(6\pi)_0-tcos \ t\ dt+[(t^2)/(2)-2cos \ t]\limits^(6\pi)_0\\\\=-I+177.65`

#Integrating I by parts:

`I=\int\limits^(6\pi)_0 tcos \ t \ dt\\\\=[\int \ tcos \ t \ dt]\limits^(6\pi)_0\\\\=[t\int cos \ t \ dt-\int ((dt)/(dt)\intcos \ t \ dt)dt]\limits^(6\pi)_0\\\\=[tsin\ t -\int sin\ t \ dt]\limits^(6\pi)_0\\\\=0`

`W=0+177.65`

Hence, work done is 177.65