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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2020
  • Medium
  • Numerical

Consider a force \(\overrightarrow{\mathrm{F}}=-\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}\). The work done by this force in moving a particle from point \(\mathrm{A}(1,0)\) to \(\mathrm{B}(0,1)\) along the line segment is  (all quantities are in \(SI\) units)

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Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(\mathrm{W}=\int_{\vec{r_i}}^{\vec{r_g}} \overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

\(\mathrm{W}=\int_{1}^{0}-\mathrm{x} \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{ydy}\)

\(W=\left.\frac{-x^{2}}{2}\right|_{1} ^{0}+\left.\frac{y^{2}}{2}\right|_{0} ^{1}\)

\(=-\left(\frac{0^{2}}{2}-\frac{1^{2}}{2}\right)+\left(\frac{1^{2}}{2}-\frac{0^{2}}{2}\right)\)

\(\mathrm{W}=1 \mathrm{J}\)

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