Question

A block of mass 5 kg initially at rest at the origin is acted on by a force along the positive X - direction represented by F= (20 +5x) N. Calculate the work done by the force during the displacement of the block from x=0 at x = 4m ​

Answer 1

Question:

• A block of mass 5 kg initially at rest at the origin is acted on by a force along the positive X - direction represented by F= (20 +5x) N. Calculate the work done by the force during the displacement of the block from x=0 at x = 4m ​

Answer:

• The work done by the body is 10 Joules

Explanation :

Given that :

• A block of mass 5 kg initially is at rest
• A force of F = ( 20 + 5x ) N is acted on it along the positive x direction along the displacement of x = 0 to x = 4m

To Find :

• The work done on the body

Formula Used :

$\bigstar \; {\underline{\boxed{ {\bf Work \; done}= \int \vec{\bf F . } \vec{\bf dx} }}}$

Required Solution :

• Since a variable force acts upon the body

$\bigstar$ Formula to find the work done

$\leadsto \; {\pink{\boxed{ {\bf Work \; done}= \int\vec{\bf F . } \vec{\bf dx} }}}$  

• Now let's find out the dot product

$\longrightarrow \vec{\sf F} . \vec { \sf dr} = \sf | F | |dx| \cos(\emptyset )$

$\bigstar$ Here ,

• As the force and the displacement are in the same direction the angle between them is 0°  [ Cos ( 0 ) = 1 ]

$\longrightarrow \rm Work \; done = \displaystyle \int^4_0 F . dx$

$\longrightarrow \rm Work \; done = \displaystyle \int^4_0 ( 20+ 5x ) \;dx$

$\longrightarrow \rm Work \; done = \displaystyle \int^4_0 20 \; dx + \int^4_0 5x \; dx$

$\longrightarrow \rm Work \; done = \displaystyle 20 \int^4_0 dx + 5 \int^4_0 x \; dx$

$\bigstar$ Integrating further

$\longrightarrow \rm Work \; done = 20 \bigg[ x \bigg]^4_0 + 5 \bigg[ \dfrac{x^2}{2} \bigg]^4_0$

$\longrightarrow \rm Work \; done = 20 \bigg[ 4 - 0 \bigg]+ 5 \bigg[ \dfrac{4^2- 0^2}{2} \bigg]$

$\bigstar$ Calculating the work done

$\longrightarrow \rm Work \; done = 20 ( 4) + 5 ( 8 - 0 )$

$\longrightarrow \rm Work \; done = 20 ( 4) + 5 ( 8 )$

$\longrightarrow \rm Work \; done = 80 + 40$

$\longrightarrow {\red{\underline{\underline{\rm{ Work \; done_{ (on \; the \; particle)}= 120 \; Joules }}}}}$

Therefore :

• The work done on the particle is 120 Joules

$\rule{400}{2}$