Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A force \mathrm{\mathrm{F}=\left(5+3 y^2\right)} acts on a particle in the \mathrm{y}-direction, where \mathrm{F} is in newton and \mathrm{y} is in meter. The work done by the force during a displacement from \mathrm{y=2 \mathrm{~m} \: to\: y=5 \mathrm{~m}\: is \: }_______________\mathrm{J}
 

Option: 1

132 J


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given :
\mathrm{F=\left(5+3 y^2\right)} in the \mathrm{ y} direction
Work done is given by
\mathrm{W}=\int_{\mathrm{y}_1}^{\mathrm{y}_2} \text { F.dy }
\mathrm{y}_1=2 \mathrm{~m}, \quad \mathrm{y}_2=5 \mathrm{~m}
\mathrm{~W}=\int_2^5\left(5+3 \mathrm{y}^2\right) \mathrm{dy}
\mathrm{W}=\int_2^5 5 \mathrm{dy}+\int_2^5 3 \mathrm{y}^2 \mathrm{dy}
\mathrm{W}=[5 \mathrm{y}]_2^5+\left[\frac{3 \mathrm{y}^3}{3}\right]_2^5
\mathrm{~W}=(5 \times 5-5 \times 2)+(125-8)
\mathrm{W}=(25-10)+117
\mathrm{W}=132 \mathrm{Joule}
 

Posted by

Gautam harsolia

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26
anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year

Latest Question