JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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)
- A \(1.5\)
- B \(1\)
- C \(2\)
- D \(0.5\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{W}=\int_{\vec{r_i}}^{\vec{r_g}} \overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\) \(\mathrm{W}=\int_{1}^{0}-\mathrm{x} \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{ydy}\)…
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