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Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the depth of the thumbtack in the bulletin board is x centimeters is given by F(x) = 1000 (1 + 2x)2 for 0 ≤ x ≤ 1 5 . Find the work W done by pushing the thumbtack all the way into the board.

Respuesta :

klefenz

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

[tex]W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx[/tex]

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

lhmarianateixeira

In this exercise, we have to use work knowledge to calculate the total, in this way:

[tex]W = 290.7[/tex]

Given some information about the exercise we find that:

  • d = 1/5 cm = 0.2 cm
  • [tex]F(x) = 1000 * (1 + 2*x)^2[/tex]

Using the working definition we find that:

[tex]W = F * d\\W = [1000*x + 2000*x^2 + 1333*X^3]\\W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3\\W = 200 + 80 + 10.7 = 290.7[/tex]

See more about work at brainly.com/question/1374468

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