AP EAMCET · PHYSICS · Work Power Energy
A particle moves in the \(x-y\) plane under the action of a force,
\(\mathbf{F}=K\left[\frac{x}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{i}}+\frac{y}{\left(x^2+y^2\right)^{\frac{3}{2}}} \hat{\mathbf{j}}\right]\) where, \(K\) is a constant. Work done by the force when the particle moves from \((0, a)\) to \((a, 0)\) along a circular path of radius \(a\) about the origin is
- A \(\frac{2 K \pi}{a}\)
- B \(\frac{K \pi}{a}\)
- C \(\frac{K \pi}{2 a}\)
- D 0
Answer & Solution
Correct Answer
(D) 0
Step-by-step Solution
Detailed explanation
According to the question, Force on a particle moves in the \(x\) - \(y\) plane, \[ \mathbf{F}=k\left[\frac{x}{\left(x^2+y^2\right)^{3 / 2}} \hat{\mathbf{i}}+\frac{y}{\left(x^2+y^2\right)^{3 / 2}} \hat{\mathbf{j}}\right] \] Now, Putting,…
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