A position dependent force, F=(7-2 x+3 x2) 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

Question

A position dependent force, F=(7-2 x+3 x2) 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

Tardigrade Admin · Accepted Answer

Force (F)=7-2 x+3 x2 ; Mass (m)=2 kg and displacement (d)=5 m. Therefore work done (W)=∫ F d x=∫ limits05(7-2 x+3 x2) d x=(7 x-x2+x3)05 =(7 × 5)-(5)2+(5)3=35-25+125=135 J

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

135

270

35

70

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

Q. A position dependent force, F=(7−2x+3x2)N acts on a small body of mass 2kg and displaces it from x=0 to x=5m. The work done in joule is

135

270

35

70

Force (F)=7−2x+3x2; Mass (m)=2kg and displacement (d)=5m. Therefore work done (W)=∫Fdx=0∫5​(7−2x+3x2)dx=(7x−x2+x3)05​ =(7×5)−(5)2+(5)3=35−25+125=135J

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

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