Question

A particle is moving along the x axis under the influence of a conservative force given by F = [−(4.00 N/m)x + (9.00 N/m2)x2]î. If U(x) = 0 when x = 0, determine the following as the particle travels from xi = 1.20 m to xf = 4.10 m.
(a) The potential energy function U(x).

(b) Change in potential energy of the particle.

(c) Change in kinetic energy of the particle.

(d) Work done on the particle by the force

Answer 1

Answer:

a)    U = 2 (x²) - 3 (x³)

b) U = - 170,839 J

c)    ΔK = 341,678 J

d)   W = 170,839 J

Explanation:

Force and potential energy are related by the equation

         F = - dU / dx

a) Let's look for potential energy

    dU = - F dx

We integrate

     dU = - ∫ (-4x +9 x²) dx

     U- Uo = (4 x² / 2 - 9 x³ / 3)

Where they indicate that U₀ = 0 for x = 0.

     U = 2 (x²) - 3 (x³)

b) We integrate and evaluate for the points of interest. Let's evaluate between the lower limit x₁ = 1.20 m and the upper boundary x₂ = 4.10 m

     U = 2 (4.1² - 1.2²) - 3 (4.1³ -1.2³)

     U = 30.74 - 201.579

     U = - 170,839 J

d) work is defined by

           W = ∫ F. dx

Since the force is in the x direction the scale product is reduced to the algebraic product

      W = ∫ (-4 x + 9 x²) dx

       W = - 2 x² + 3 x³

We evaluate between the limits

      W = -2 (4.1² -1.2²) + 3 (4.1³- 1.2³)

      W = -30.74 + 201.579

      W = 170,839 J

c) let's look for the kinetic energy of the particle

       W = ΔK + ΔU

       ΔK = W - ΔU

       ΔK = 170,839 - (-170,839)

       ΔK = 341,678 J

Answer 2

Answer:

(a) ux = 2x² -3x³

(b) Change in potential energy = -170.839j

(c) Change in kinetic energy = 170.839j

(d) Workdone by force = 170.839j

Explanation:

See the attached for the calculation