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=−1Nm−1. The work done on the particle by this force F→ will be n×0.25 Joule.Find the value of n

For AB:y= const F→=−yi^−2xj^⇒F→=−i^−2xj^dr→=dxi^ So, WAB=∫x=01 F→⋅dr→=∫01 −dx=−1J For BC:x=constF→=−yi^−2j^dr→=dyjWBC=∫y=10.5 −2dy⇒WBC=+1J For CD:y=const F→=−0.5i^−2xj and dr→=dxi^WCD=∫10.5 −0.5dx=0.25J For DE:x= const F→=−yi^−j^dr→=dyj^WDE=∫y=0.50 −dy=+0.5J For EF:y=0 F→=−2×j^dr→=dxi^⇒WEF=0 For FA:x=0 F→=−yi^dr→=dyj^⇒WFA=0WNet=−1+1+0.25+0.5=0.75J

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 $α = - 1 N m^{- 1}$ . The work done on the particle by this force $\vec{F}$ will be $n \times 0.25$ Joule.Find the value of n

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} For A B : y = const \\ \vec{F} = - y \hat{i} - 2 x \hat{j} \\ \Rightarrow \vec{F} = - \hat{i} - 2 x \hat{j} \\ \vec{d r} = d x \hat{i} \\ So, W_{A B} = \int_{x = 0}^{1} \vec{F} \cdot \vec{d r} = \int_{0}^{1} - d x = - 1 J \end{array}$

$\begin{array}{l} For BC : x = const \\ \vec{F} = - y \hat{i} - 2 \hat{j} \\ \vec{d r} = d y j \\ W_{B C} = \int_{y = 1}^{0.5} - 2 d y \\ \Rightarrow W_{B C} = + 1 J \end{array}$

$\begin{array}{l} For CD:y=const \\ \vec{F} = - 0.5 \hat{i} - 2 x j and d \vec{r} = d x \hat{i} \\ W_{C D} = \int_{1}^{0.5} - 0.5 d x = 0.25 J \end{array}$

$\begin{array}{l} For DE : x = const \\ \vec{F} = - y \hat{i} - \hat{j} \\ \vec{d r} = d y \hat{j} \\ W_{D E} = \int_{y = 0.5}^{0} - d y \\ = + 0.5 J \end{array}$

$\begin{array}{l} For EF : y = 0 \\ \vec{F} = - 2 \times \hat{j} \\ d \vec{r} = d x \hat{i} \\ \Rightarrow W_{E F} = 0 \\ For F A : x = 0 \\ \vec{F} = - y \hat{i} \\ d \vec{r} = d y \hat{j} \\ \Rightarrow W_{F A} = 0 \\ W_{Net} = - 1 + 1 + 0.25 + 0.5 = 0.75 J \end{array}$