A single conservative force acts on a 4.50-kg particle within a system due to it

Question

A single conservative force acts on a 4.50-kg particle within a system due to its interaction with the rest of the system. The equation

Fx = 2x + 4

describes the force, whereFx is in newtons and x is in meters. As the particle moves along the x axis from x = 0.98 m to x = 4.55 m, calculate the following.

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

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

(c) the kinetic energy the particle has at x = 4.55 m if its speed is 3.00 m/s at x = 0.98 m? (J)

Explanation / Answer

a) work done = integral of (F.dx) from 0.98 m to 4.55 m

W = integral of (2x +4).dx from 0.98 m to 4.55 m

W = [x^2 + 4x ] from 0.98 m to 4.55 m

W = [4.55^2 + 4*4.55] - [ 0.98^2 + 4*0.98] =34.02 J

b) change in P.E> = - work done

= - 34.02 J

c) net work done = change in k.E.

34.02 = 4.50v^2 /2 - 4.50 x 3^2 /2

v = 4.91 m/s

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