Q130 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Physics

A horizontal force $\mathbf{F}=\left(g-x^2\right) \hat{\mathbf{i}} \mathrm{N}$ acts on a wooden block resting on a horizontal smooth surface. The work done to move the block from $x=0$ to $x=3 \mathrm{~m}$ (in $\mathrm{J}$ ) is (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )

  • A
    24
  • B
    35
  • C
    30
  • Correct answer
    D
    21

Solution

Given that, force acting on wooden block $$ \mathbf{F}=\left(g-x^2\right) \hat{\mathbf{i}} \mathrm{N} $$ Displacement, $d \mathbf{x}=d x \hat{\mathbf{i}}$ Work done by $\mathbf{F}$ for displacement from $x=0$ to $x=3 \mathrm{~m}$ $$ \begin{aligned} W & =\int_0^3 \mathbf{F} \cdot d \mathbf{x}=\int_0^3\left(g-x^2\right) d x=\left[g x-\frac{x^3}{3}\right]_0^3 \\ & =\left(3 g-\frac{27}{3}\right)-0=3 \times 10-9 \\ & =30-9=21 \mathrm{~J} \end{aligned} $$

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

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