A net external force is applied to a 6.90-kg object that is initially at rest.?

Question

A net external force is applied to a 6.90-kg object that is initially at rest. The net force component along the displacement of the object varies with the magnitude of the displacement as shown in the drawing.

(b) What is the speed of the object at s = 20.0 m?

Here's the picture

http://i143.photobucket.com/albums/r132/carlarnold1/wevb.jpg

Answer 1

During the first 10 m the force increases in a linear fashion. Because it's linear, you can average it for the first 10m: 5 N. This easy way will give the same velocity at the end of the 10 m as if you did it by area under the curve. According to Newton's 2nd law, this gives the object acceleration a1:
F = m*a1 where F = 5 N and m = 6.90 kg.

Use a1 in the equation V^2 = Vo^2 + 2*a1*10 m where Vo = 0. Solve for V.

For the second 10 m section, the force is steady at 10 N. Again use Newton's 2nd law to get a2:
F = m*a2 where F = 10 N and m = 6.90 kg.

Use a2 in the equation V^2 = Vo^2 + 2*a2*10 m where Vo = V from the last part. Solve for V.

Answer 2

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