Q.
The relationship between the force $F$ and position $x$ of a body is as shown in figure. The work done in displacing the body from $x = 1 \, m $ to $x = 5 \, m$ will be :
BITSATBITSAT 2005
Solution:
Work done = area enclosed by $F-x$ graph
= area of $ABNM$ + area of $CDEN$ - area of $EFGH$ + area of $HIJ$
$ = 1 \times 10 + 1 \times 5 + \frac{1}{2} \times 1 \times 10$
$ = 10 + 5 - 5 + 5 = 15\, J $