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A CD case slides along a floor in the positive direction of an x axis while an applied force a acts on the case. The force is directed along the x axis and has the x component Fax = 8.0x – 5.0x2, with x in meters and Fax in newtons. The case starts at rest at the position x = 0, and it moves until it is again at rest.

(a) At what position is the work maximum, and

(b) what is that maximum value?

(c) At what position has the work decreased to zero?

(d) At what position is the case again at rest?

Respuesta :

cjmejiab

To solve this problem we will use the work theorem which describes the change of force in proportion to the distance traveled.

Our function is given as:

[tex]F_{ax} = 8x-5x^2[/tex]

PART A) The maximum work is given in the position in which the function of the Force is equal to zero, therefore

[tex]F_{ax} = 0[/tex]

[tex]8x-5x^2 = 0[/tex]

[tex]8= 5x[/tex]

[tex]x = 1.6m[/tex]

PART B) The maximum work done is given from the 0 position to the maximum work distance, therefore:

[tex]W = \int^{1.6}_0 (8x-5x^2)dx[/tex]

[tex]W = \Big[\frac{8x^2}{2}-\frac{5x^3}{3}\Big]^{1.6}_0[/tex]

[tex]W = \frac{8(1.6^2)}{2}-\frac{5(1.6^3)}{3}[/tex]

[tex]W = 3.44J[/tex]

PART C) At the moment when the work reaches zero, we will simply match the integral previously found to zero value so

[tex]w = \frac{8x^2}{2}-\frac{5x^3}{3}[/tex]

[tex]\frac{8x^2}{2}=\frac{5x^3}{3}[/tex]

[tex]24=10x[/tex]

[tex]x = 2.4m[/tex]

PART D ) If the work is zero in position 2.4, do not open movement, as the distance to travel will also be 0. The body will be at rest again 2.4m from its initial position.

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