A 5 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. Find the work done by this force as the block moves from the origin to x = 8m.

Explanation

Work = int F. Dx for a variable force. So the work from x = 0 to x = 8 is the area under the F-x curve. Here area below the x - axis is considered negative.
`W = (10 xx 2) + 1/2 (10)(4-2) + 0+ 1/2 (-5)(8-6)`
` = 25 J`