Q110 / 200 · NEET 2005 · Physics

A force $F$ acting on an object varies with distance $x$ as shown here. The force is in $N$ and $x$ in $\mathrm{m}$. The work done by the force in moving the object $x=0$ to $x=6 \mathrm{~m}$ is: Figure

  • A
    $18.0 \mathrm{~J}$
  • Correct answer
    B
    $13.5 \mathrm{~J}$
  • C
    $9.0 \mathrm{~J}$
  • D
    $4.5 \mathrm{~J}$

Solution

Work done $=$ Area under $=f-x$ curve. $$ \begin{aligned} & =\text { area of trapezium }=\frac{1}{2} \times(6+3) \times 3 \\ & =\frac{27}{2}=13.5 \mathrm{~J} \end{aligned} $$

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

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