Question

Fvector = x²i+2yj , and body is displaced from (1,2) to (3,4) . Find the work done.

Answer 1

Given:

Force $\rm (\overrightarrow{F})$ = $\rm x^2\hat{i} + 2y\hat{j}$

Initial position $\rm (r_i)$ = (1,2)

Final position $\rm (r_f)$ = (3,4)

To Find:

Work Done (W)

Answer:

For a varying force the work done can be expressed as a definite integral of force over displacement.

$\bf \leadsto W = \displaystyle \rm \int\limits_{r_i}^{r_f} \overrightarrow{F}. \overrightarrow{dr}\\ \\ \rm \leadsto W = \displaystyle \rm \int\limits_{(1,2)}^{(3,4)} ( x^2\hat{i} + 2y\hat{j}). (dx \hat{i} + dy \hat{j}) \\ \\ \rm \leadsto W = \displaystyle \rm \int\limits_{1}^{3} x^2 .dx + \displaystyle \rm \int\limits_{2}^{4} 2y.dy \\ \\ \rm \leadsto W = \dfrac{ {x}^{3} }{3} \Bigg|_{1}^{3} + {y}^{2} \bigg|_{2}^{4} \\ \\ \rm \leadsto W = \dfrac{ ({3}^{3} - {1}^{3}) }{3} + ( {4}^{2} - {2}^{2} ) \\ \\ \rm \leadsto W = \dfrac{ (27 - 1) }{3} + ( 16 - 4 ) \\ \\ \rm \leadsto W = \dfrac{ 26}{3} + 12 \\ \\ \rm \leadsto W = \dfrac{ 26 + 36}{3} \\ \\ \rm \leadsto W = \dfrac{ 62}{3} \: J \\ \\ \rm \leadsto W = 20.67 \: J$

$\therefore$ Work Done (W) = 20.67 J

Answer 2

W = ( x²i + 2yj ) × ( dxi + dyj )

= ( x² × dx ) + ( 2y × dy )

= x / 3 ( 3 , 1 ) + y² ( 4 , 2 )

= [ ( 3 × 3 × 3 - 1 × 1 × 1 ) / 3 ] + [ 4 × 4 - 2 × 2 ]

= ( 27 - 1 ) / 3 + ( 16 - 4 )

= ( 26 / 3 ) + 12

LCM = 3

= 26 + 36 / 3

= 62 / 3

= 20.6