Q.
The force $F$ and position $x$ of a body relates as shown in figure. What will the work done in displacing the body from $x=1 \, m$ to $x=5 \, m$ be ?
NTA AbhyasNTA Abhyas 2020
Solution:
$work$ done equal to area under $F-x$ graph.
Since $work$ is a scalar quantity, so when $area$ is positive $work$ is also positive and vice versa.
So
$W$ $=$ area of rectangle $abce$ $+$ area of rectangle $-$ area of rectangle $ghij+$ area of triangle $jkl$
$W=\left(2 - 1\right)\times \left(10 - 0\right)+\left(3 - 2\right)\times \left(5 - 0\right)-\left(4 - 3\right)\times \left(5 - 0\right)+\frac{1}{2}\times \left(5 - 4\right)\times \left(10 - 0\right)\Rightarrow W=1\times 10+1\times 5-1\times 5+\frac{1}{2}\times 1\times 10\Rightarrow W=10+5-5+5\Rightarrow W=15J$
$work$ done equal to area under $F-x$ graph.
So
$W$ $=$ area of rectangle $abce$ $+$ area of rectangle $-$ area of rectangle $ghij+$ area of triangle $jkl$
$W=\left(2 - 1\right)\times \left(10 - 0\right)+\left(3 - 2\right)\times \left(5 - 0\right)-\left(4 - 3\right)\times \left(5 - 0\right)+\frac{1}{2}\times \left(5 - 4\right)\times \left(10 - 0\right)\Rightarrow W=1\times 10+1\times 5-1\times 5+\frac{1}{2}\times 1\times 10\Rightarrow W=10+5-5+5\Rightarrow W=15J$