A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 1.40 m to x = 5.80 m, calculate the following.

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

(b) the change in the potential energy of the system

(c) the kinetic energy the particle has at x = 5.80 m if its speed is 3.00 m/s at x = 1.40 m

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  1. (a) The work done is the integral of F dx from 1.4 to 5.8.

    If you do not know how to integrate, multiply the AVERAGE force by 4.4 m. The average force is in this case the value at x = 3.60 m

    (b) The P.E. change is MINUS the work done in part (a). Just change the sign.

    (c) To get the new value of kinetic energy, add the work done to (1/2) M V^2, where Vo = 3 m/s
    (K.E.)final = (Work done)+ (M/2)Vo^2

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  2. Thanks so much for your help drwls!

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