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JEE Advanced · Physics · 6. Work Power Energy

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force F=αyi^+2αxj^N, where x and y are in meter and α=-1 N/m-1. The work done on the particle by this force F will be ____ Joule.

  1. A 0.25
  2. B 0.5
  3. C 0.1
  4. D 0.75
Verified Solution

Answer & Solution

Correct Answer

(D) 0.75

Step-by-step Solution

Detailed explanation

dw=Fdr
dw=aydx+2axdy
Now, total work done in whole path is given by,
\(W=W_{A B}+W_{B C}+W_{C D}+W_{D E}+W_{E F}+\)\(W_{F A}\)
\(A \rightarrow B, y=1, d y=0, W_{A \rightarrow B}=\int \alpha y d x=\alpha \cdot\) \(1 \int_0^1 d x=\alpha\)
\(B \rightarrow C, x=1, d x=0, W_{B \rightarrow C}=2 \alpha \cdot 1 \int_1^{0.5}\) \(d y=-2 \alpha(0.5)=-\alpha\)
\(C \rightarrow D, y=0.5, d y=0, W_{C \rightarrow D}=\int_1^{0.5} \alpha y d x\)\(=\alpha \cdot \frac{1}{2} \int_1^{0.5} d x=-\frac{\alpha}{4}\)
\(D \rightarrow E, x=0.5, d x=0, W_{D \rightarrow E}=2 \alpha \int x d y=\)\(2 \alpha \cdot \frac{1}{2} \int_1^{0.5} d y=-\frac{\alpha}{2}\)
\(E \rightarrow F, y=0, d y=0, W_{E F}=0\)
\(F \rightarrow A, x=0, d x=0, W_{E \rightarrow A}=0\)
\(\therefore W=\alpha-\alpha-\frac{\alpha}{4}-\frac{\alpha}{2}=-\frac{3 \alpha}{4}\)
Given \(\alpha=-1 \Rightarrow W=+\frac{3}{4} J=0.75 J\)
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