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A variable force of 5x^-2 pounds moves an object along a straight line from the origin. Calculate the work done.

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 positive 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.

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