Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A particle is moved along the x-axis by a force that measures 9/(6 x) pounds at a point x feet from the origin. Find the work W done in moving the particle from the origin to a distance of 12 ft.

Answer 1

Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

$F = \dfrac{9}{6^x}$

This force is not a constant force. For a non-constant force, the work done, W, is

$W = \int\limits^{x_2}_{x_1} {F(x)} \, dx$

with $x_1$ and $x_2$ the initial and final displacements respectively.

From the question, $x_1 =0$ and $x_2 = 12$.

Then

$W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx$

Evaluating the indefinite integral,

$\int\limits \dfrac{9}{6^x} \, dx =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx$

From the rules of integration,

$\int\limits a^x\, dx = \dfrac{a^x}{\ln a}$

$9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x$

Returning the limits,

$\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}$