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4) A single conservative force acts on a 5.00 kg particle within

a system due to its interaction with the rest of the system. The

equation Fx = 2x 4 describes the force, where Fx is in newtons

and x is in meters. As the particle moves along the x axis from x =

1.00 m to x = 5.00 m, calculate

a) the work done by this force on the particle,

b) the change in the potential energy of the system, and

c) the kinetic energy the particle has at x = 5.00 m if its speed

is 3.00 m/s at x = 1.00 m.  Edit Comment  Solutions
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Fdx = (2x 4)dx

integrated w/r/t x yields

W = x^2 4x

evaluated from 1.0 to 5.0 m:

W = 5.0^2 4*5.0 - 1^2 - 4*1 = 40 J

b) -40 J (force is conservative)

c) dU = -dK (change in potential = change in kinetic)

Ki = 0.5*m*v^2 = 0.5*5kg*(3m/s)^2 = 22.5J

Kf = Ki dK = 22.5J 40J = 62.5 J  Edit Comment  Close Choose An Image
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