A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force vec F =(α y hat i +2 α x hat j ) N , where x and y are in meter and α=-1 Nm -1 . The work done on the particle by this force vec F will be \ \ \ \ Joule.

Physics · Work, Power & Energy

JEE Advanced 2019 — Paper 1 — Question 13

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force $\vec{F}=(\alpha y \hat{i}+2 \alpha x \hat{j}) N$ , where x and y are in meter and $\alpha=-1 \mathrm{Nm}^{-1}$ . The work done on the particle by this force $\vec{F}$ will be $\_\_\_\_$ Joule.

Answer: 0.75

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{w}_{\mathrm{AB}}=\int_{0}^{1} \alpha y d x=-1$

$\mathrm{w}_{\mathrm{BC}}=\int_{1}^{0.5} 2 \alpha \mathrm{xdy}=+1$

$\mathrm{w}_{\mathrm{CD}}=\int_{1}^{0.5} \alpha y d x=+0.25$

$\mathrm{w}_{\mathrm{DE}}=\int_{0.5}^{0} 2 \alpha \mathrm{xdy}=+0.5$

$\mathrm{w}_{\mathrm{EF}}=\mathrm{W}_{\mathrm{FA}}=0$

$\mathrm{W}_{\text {net }}=0.75$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Work, Power & Energy
Topic: Work Done by a Constant and Variable Force