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

135

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270

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35

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70

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Solution

The correct option is A $135$
Work done by variable force
$W = \int_{0}^{5} F d x = \int_{0}^{5} (7 - 2 x + 3 x^{2}) . d x$

$W = {[7 x - \frac{2 x^{2}}{2} + \frac{3 x^{3}}{3}]}_{0}^{5}$

$W = 135 J$

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

The correct option is A 135Work done by variable force W=∫50Fdx=∫50(7−2x+3x2).dx W=[7x−2x22+3x33]50 W=135 J

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

The correct option is A $135$
Work done by variable force
$W = \int_{0}^{5} F d x = \int_{0}^{5} (7 - 2 x + 3 x^{2}) . d x$

$W = {[7 x - \frac{2 x^{2}}{2} + \frac{3 x^{3}}{3}]}_{0}^{5}$

$W = 135 J$