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)

Pick an option to check your answer.
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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