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

Answer:

177.65

Explanation:

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

$W=\int\limits_C F.dr$

Given that:

$r(t) = (t -sin(t)i + (1 -cos(t)j,\\\\\frac{dr}{dt}=(1-cos t)i+sin t\ j\\\\dr=(1-cos\ t)i +sin\ t \ j)dt\\\\F(x,y)=x\ i +(y+6)j\\\\F(r(t))=(t-sin \ t) i+((1-cos \t)+2)j$

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

#Integrating I by parts:

$I=\int\limits^{6\pi}_0 tcos \ t \ dt\\\\=[\int \ tcos \ t \ dt]\limits^{6\pi}_0\\\\=[t\int cos \ t \ dt-\int (\frac{dt}{dt}\intcos \ t \ dt)dt]\limits^{6\pi}_0\\\\=[tsin\ t -\int sin\ t \ dt]\limits^{6\pi}_0\\\\=0$

$W=0+177.65$

Hence, work done is 177.65