A single conservative force acts on a 5.10-kg particle within a system due to it

Question

A single conservative force acts on a 5.10-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 = 1.08 m to x = 7.30 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 = 7.30 m if its speed is 3.00 m/s at x = 1.08 m

Explanation / Answer

Given that -

Fx = 2x + 4

So,

Fdx = (2x+4)dx
integrate the above with respect to x and t, we have -

W = x^2 + 4x

(a) Work done from x = 1.08 m to x = 7.30 m

W = 7.30^2 + 4*7.30 - 1.08^2 - 4*1.08 = 53.29 + 29.2 - 1.17 - 4.32 = 77.0 J

(b) Change in potential energy of the particle = -77.0 J (force is conservative)

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

Ki = (1/2)*m*v^2 = 0.5*5.10kg*(3m/s)^2 = 22.95 J
Kf = Ki + dK = 22.95 J + 77.0 J = 99.95 J

So, kinetic energy of the particle at x = 7.30 m is 99.95 J.

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