Question

A block of mass 5 kg initially at rest at the origin is acted on by a force along the X-
positive direction represented by F = (20+5x) N.Calculate the work done by the
force during the displacement of the block from x = 0 to x = 4 m​

Answer 1

☆whenever variable force acts...integration is required

W=Fds(variable force)

$w = (20 + 5x)dx$

•

$w = 20dx + 5xdx$

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$w = 20x + 5 \frac{ {x}^{2} }{2}$

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$w = 20(4 - 0) + \frac{5}{2} ( {4}^{2} - {0}^{2} )$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: w = 20 \times 4 + \frac{5}{2} \times 16 \\ w = 80 + 40 \\ w = 120joules$

so..net work done is +120J