This question aims to find the **work done** in moving the object within a **certain distance** when a **variable force** of $ 5x ^ {-2 } $ acts on the object.

**Work is done** by displacing a body when a certain force is applied to it. It is represented by $ W = F \times d $, where **F** is the **force acting** on the body, **d** is the **displacement,** and **W** is the **work done** on the body.

We can split the force into **two components,** also called the **resolution of force,** to get an idea of the direction of the force. The two components of force are the **horizontal** component and the **vertical component**. The horizontal component of force acts along the **x-axis** and the vertical component of the force acts along the **y-axis.**

They are represented by:

\[ F _ x = F cos \theta \]

\[ F _ y = F sin \theta \]

## Expert Answer

An object moves when a force is applied along the **x-axis** in the p**ositive direction** from a certain distance **x = a** to **x = b** and a then this force becomes the function **f (x)**. The work done on this force is given by:

\[ W = \int_{ a }^{ b } f ( x ) \,dx \]

When an object moves x units from its origin along a **straight line** in such a way that the initial **x is 1** and the final value of **x is 10,** then the expression will become:

$ f ( x ) = 5 x ^ { -2 } $ and limits are $ [ a , b ] = [ 1 , 10 ] $

Putting values in the above expression:

\[ W = \int_{ 1 }^{ 10 } 5 x ^ { – 2 } \,dx \]

By applying the power rule of integration:

\[ W = \Bigr[ \frac { 5 x ^ { – 2 + 1 }} { – 2 + 1 } \Bigr] _ { 1 }^{ 10 } \]

\[ W = \Bigr[ \frac { 5 x ^ { – 1 }} { – 1 } \Bigr] _ { 1 }^{ 10 } \]

\[ W = \Bigr[ \frac { – 5 } { x } \Bigr] _ { 1 }^{ 10 } \]

\[ W = \Bigr[ \frac { – 5 } { 10 } \Bigr]- \Bigr[ \frac { – 5 } { 1 } \Bigr] \]

\[ W = – 0 . 5 + 5 \]

\[ W = 4 . 5 lb . ft \]

## Numerical Solution

**The work done along the horizontal direction is $ 4 . 5 lb. ft $.**

## Example

Find **work done** along the positive **x-direction** when **force F** is acting on the body and displaces it from **x = 1** to **x = 8**.

\[ W = \int_{ 1 }^{ 10 } 5 x ^ { – 2 } \,dx \]

By applying the power rule of integration:

\[ W = \Bigr[ \frac{5x^{-2+1}}{-2+ 1 } \Bigr]_ { 1 }^{ 8 } \]

\[ W = \Bigr[\frac{5x^{-1}}{-1}\Bigr] _ { 1 }^{ 8 }\]

\[ W = \Bigr[\frac{-5}{x}\Bigr] _ {1}^{8}\]

\[ W = \Bigr[\frac{-5}{8}\Bigr] – \Bigr[\frac {-5}{1}\Bigr]\]

\[ W = -0.625 + 5 \]

\[ W = 4 . 375 lb . ft \]

*Image/Mathematical drawings are created in Geogebra**.*