Question

A particle is moved along the x-axis by a force that measures 10/(1 + x)2 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 9 ft.

Answer 1

Answer:

Work done, W = 12.20 J

Explanation:

It is given that,

Force acting on the particle, $F=\dfrac{10}{(1+x)^2}$

Where

x is the distance in feet from the origin

We need to find the work done by the particle from the origin to a distance of 9 ft.

$W=\int\limits^9_0 {F} \, dx$

$W=\int\limits^9_0 {\dfrac{10}{(1+x)^2}} \, dx$

After solving the above integration using calculator we get,

W = 9 Pound-ft

On converting 9 pound-ft in J, we get the work done as, W = 12.20 J. Hence, this is the required solution.

Answer 2

The work done in moving particles is defined as the product of force and displacement.The particle is moved along the x-axis is from 0 to 9 feet then work done is 9 pound-feet.

What is work done?

The work done in moving particles is defined as the product of force and displacement. It can also find with the help of integration by integrating force with respect to the displacement.

Given,

The force acting on the particle is $F=\dfrac{10}{(1+x)^2}$

The distance of the particle from the origin = 9 pounds

We know the formula for the work done is given by

The work done on the particle will be W= ∫F.dx

$\rm W= 10 \int\limits^9_0 {} \, \frac{dx}{(1+x)^{2} }\\\\W = 10 [\dfrac{1}{-(x+1)}]^9_0\\\\W = 10[-0.1 + 1]\\\\W = 10[0.9]\\\\W = 9$

Thus, the work done is 9 pound-feet.

To learn more about the work done refer to the link.

https://brainly.com/question/3902440