Question

Evaluate the amount of work done by the force field F(x,y)=1x2i+yexjF(x,y)=1x2i+yexj on a particle that moves along the curve C:x=y2+1C:x=y2+1 from (1,0)(1,0) to (2,1)(2,1).

Answer 1

Answer:

$\frac{7}{3} + \frac{e^2 - e}{2}$

Step-by-step explanation:

By definition:

Work done along the path is the line integral along that path denoted as:

Work Done = $\int\limits^C {F} \, dr$

Note:  dr = dx i + dy j

Given that: $F (x,y) = x^2 i + ye^x j$

F (x, y) dot product with dr = $x^2 dx + ye^x dy$

Work done = $\int\limits^C {(x^2 dx + ye^x dy)}$  ... Eq 1

Given that C: $y = \sqrt{x-1}$

$dy = \frac{dx}{2\sqrt{x-1} }$

Replace the value of y and dy in Eq 1

$Work done = \int\limits^C ({x^2 + \frac{e^x}{2} }) \, dx$

Limits of x are 1 to 2 respectively

$Work done = \int\limits^2_1 ({x^2 + \frac{e^x}{2} }) \, dx$

= $(\frac{x^3}{3} + \frac{e^x}{2})\limits^2_1$

Evaluate limits to obtain

Work Done = $\frac{7}{3} + \frac{e^2 - e}{2}$