Question

A position dependent force, F = (7 - 2x + 3x$^2$) N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joule is

• A. 135
• B. 270
• C. 35
• D. 70

Answer 1

Answer: (A)

135

Answer 2

Force (F) = 7 - 2x + 3x$^2$; Mass (m) = 2 kg and displacement (d) = 5 m.
Therefore work done
$(W)=\int F dx = \int \limits_{0}^5(7-2x+3x^2)dx=(7x-x^2+x^3)_0^5$
=$(7 \times 5)-(5)^2 +(5)^3 = 35-25+125=135 \,J.$