A position dependent force $F = 7 - 2x + 3x^{2}$ $newton$ acts on a small body of mass $2\ kg$ and displaces it from $x = 0$ to $x = 5m$. The work done in $joules$ is

Question

Dr. Abubacker · Accepted Answer

(d) $W = \int_{0}^{5}{Fdx} = \int_{0}^{5}{\left( 7 - 2x + 3x^{2} \right)dx =}\left\lbrack 7x - x^{2} + x^{3} \right\rbrack_{0}^{5}\ $
$= 35 - 25 + 125 = 135\ J$

Q. A position dependent force $F = 7 - 2x + 3x^{2}$ $newton$ acts on a small body of mass $2\ kg$ and displaces it from $x = 0$ to $x = 5m$. The work done in $joules$ is

Explanation

Reviewed by Dr. Abubacker

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