Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A particle is acted upon by constant forces 4\hat{i}+\hat{j}-3\hat{k}\; and\; 3\hat{i}+\hat{j}-\hat{k}\; which displace it from a point \hat{i}+2\hat{j}+3\hat{k}\; to the point 5\hat{i}+4\hat{j}+\hat{k}\; . The work done in standard units by the forces is given by

  • Option 1)

    25

  • Option 2)

    30

  • Option 3)

    40

  • Option 4)

    15

 

Answers (1)

As we learnt in 

Applications of Vectors -

\vec{W}= \vec{F}.\vec{S}

- wherein

Work done against a constant force \vec{F} over a displacement \vec{S} 

 

 Work done by forces \vec{F_1}\: and \:\vec{F_2}\:is

(\vec{F_1}\: + \:\vec{F_2}).\vec{d}

\vec{F_1}\: + \:\vec{F_2}=(4\hat{i}+\hat{j}-3\hat{k}) + (3\hat{i}+\hat{j}-\hat{k})

= (7\hat{i}+\hat{2j}-\hat{4k})

\vec{d}= (5\hat{i}+\hat{4j}+\hat{k})- (\hat{i}+\hat{2j}+\hat{3k})

\vec{d}= (4\hat{i}+\hat{2j}-\hat{2k})

Work done = =(\vec{F_1}+\vec{F_2}).\vec{d}

= (7\hat{i}+\hat{2j}-\hat{4k}).(4\hat{i}+\hat{2j}-\hat{2k})

= 28+4+8 =40


Option 1)

25

Incorrect

Option 2)

30

Incorrect

Option 3)

40

Correct

Option 4)

15

Incorrect

Posted by

Sabhrant Ambastha

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 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question