First slide

Work done by Diff type of forces

Question

A position dependent force, $F = 7 - 2 x + 3 x^{2}$ newton 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

Easy

A

70
B

270
C

35
D

135

Solution

$W = \int_{0}^{5} F \times d x = \int_{0}^{5} (7 - 2 x + 3 x^{2}) d x = [7 x]_{0}^{5} - {[\frac{2 x^{2}}{2}]}_{0}^{5} + {[\frac{3 x^{3}}{3}]}_{0}^{5} = 35 - 25 + 125 - 135 joule$

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