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The force on a 3-kg object as a function of position is shown in the figure. If an object is moving at 2.50 m/s when it is lo
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Answer #1

Work done on the object between x = 2 m to x = 8 m is the area under F-x curve.
W = (1 - 0) * (8 - 2) * 0.5 * (2 - 1) * (3 - 2)
= 6 + 0.5 = 6.5 J
-------------------------------------
Initial velocity, vi = 2.5 m/s
Consider vf as the final velocity
Mass of the object, m = 3 kg
Change in kinetic energy = 1/2 m * vf2 - 1/2 * m * vi2
= 1/2 * m * (vf2 - vi2)
-------------------------------------
W = 1/2 * m * (vf2 - vi2)
6.5 = 0.5 * 3 * (vf2 - 2.52)
vf2 - 6.25 = 4.33
vf2 = 10.58
vf = SQRT[10.58]
= 3.25 m/s

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