First slide
Work
Question

A position dependent force F=72x+3x2newton acts on a small body of mass 2 kg and displaces it from x=0 to x=5m. The work done in joules is

Moderate
Solution

W=05Fdx=05(72x+3x2)dx=[7xx2+x3]05

= 35 – 25 + 125 = 135 J

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