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

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270

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135

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Solution

The correct option is D 135
$W = \int F d x$
$\int_{0}^{5} (7 - 2 x + 3 x^{2}) d x = {(7 x - \frac{2 x^{2}}{2} + \frac{3 x^{3}}{3})}_{0}^{5} = 135 J$

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A position dependent force F=7−2x+3x2N acts on a small object of mass 2kg to displace it from x=0 to x=5m. The work done in joules is-

The correct option is D 135W=∫Fdx∫50(7−2x+3x2)dx=(7x−2x22+3x33)50=135J

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

The correct option is D 135
$W = \int F d x$
$\int_{0}^{5} (7 - 2 x + 3 x^{2}) d x = {(7 x - \frac{2 x^{2}}{2} + \frac{3 x^{3}}{3})}_{0}^{5} = 135 J$