Q38 / 50 · Manipal (MET) 2023 · Physics

A force $F=(2+x)$ acts on a particle in $x$-direction, where F is in newton and $x$ in metre. The work done by this force during'a displacement from $x=1.0 \mathrm{~m}$ to $x=2.0 \mathrm{~m}$ is

  • A
    $2 J$
  • Correct answer
    B
    $3.5 J$
  • C
    $4.5 J$
  • D
    None of these

Solution

The work done in small displacement from $x$ to $x+d x$ is $d W=F d x=(2+x) d x$ Hence, $W=\int_1^2 d W=\int_1^2(2+x) d x=\int_1^2 2 d x+\int_1^2 x d x$ $=\left[2 x+\frac{x^2}{2}\right]_1^2=3.5 \mathrm{~J}$

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

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