A position dependent force `F=7-2x+3x^(2)` acts on a small body of mass 2 kg and displaced it from `x=0` to `x=5m`. Calculate the work done in joule.

Options

A. 138 J
B. 135 J
C. 136 J
D. 137 J

Answer

B

Explanation

Work done, `W=int_(x_(1))^(x_(2)) F dx=int_(0)^(5)(7-2x+3x^(2))dx`
Here, the body changes its position from `x=0` to `x=5`
`=[7 x-(2x^(2))/2+(3x^(3))/3]_(0)^(5)=[7 (5)-(5)^(2)+(5)^(2)-0]=135 J`