Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! When forces F1 ,F2 ,F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary If the force F1 is now removed then the acceleration of the particle is :

When forces F1 ,F2 ,F3 are acting on a particle of mass  m such that F2 and F3  are mutually perpendicular, then the particle remains stationary If the force F1 is now removed then the acceleration of the particle is :

  • Option 1)

    F1/ m

  • Option 2)

    F2F3  /mF1

  • Option 3)

    (F2 - F3) / m

  • Option 4)

    F2 / m

 

Answers (1)

best_answer

As we learnt in

Impulse Momentum Theorem -

\vec{F}=\frac{d\vec{p}}{dt}

\int_{t_{1}}^{t_{2}}\vec{F}dt=\int_{p_{1}}^{p_{2}}\vec{dp}

- wherein

If \bigtriangleup t  is increased, average force is decreased

 

 

 

:  F2 and F3  have a resultant equivalent to F1

\therefore  Aceleration = F1/m  .

Correct option is 1.


Option 1)

F1/ m

This is the correct option.

Option 2)

F2F3  /mF1

This is an incorrect option.

Option 3)

(F2 - F3) / m

This is an incorrect option.

Option 4)

F2 / m

This is an incorrect option.

Posted by

Plabita

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

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'
anam-khan

BTech Admissions 2025 Live: JEE Mains, SRMJEE, VITEEE dates out; AEEE, MET registrations underway

Latest Question