Q123 / 200 · NEET 2011 (Screening) · Physics

Force $F$ on a particle moving in a straight line varies with distance $d$ as shown in the figure. The work done on the particle during its displacement of $12 \mathrm{~m}$ is

A
$21 \mathrm{~J}$
B
$26 \mathrm{~J}$
Correct answer
C
$13 \mathrm{~J}$
D
$18 \mathrm{~J}$

Solution

Work done $=$ Area under $(F-x)$ graph $$ \begin{aligned} & =2 \times(7-3)+\frac{1}{2} \times 2 \times(12-7) \\ & =8+\frac{1}{2} \times 10 \\ & =8+5=13 \mathrm{~J} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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