A force`vecF=(7-2x+3x^(2))` N is applied on a 2 kg mass which displaces it from x = 0 to x = 5 m. Work done in joule is -

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
`=[7x-(2x^(2))/(2)+(3x^(3))/(3)]_(0)^(5)=[7(5)-(5)^(2)+(5)^(3)-0]=135 J`