Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A small particle moves to position 5iˆ − 2jˆ + kˆ from its initial position 2iˆ + 3jˆ − 4kˆ under the action of force 5iˆ + 2jˆ + 7kˆ N. The

A small particle moves to position 5iˆ − 2jˆ + kˆ from its initial position 2iˆ + 3jˆ − 4kˆ under the action of force 5iˆ + 2jˆ + 7kˆ N. The value of work done will be ________ J.

Option: 1

40


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \Delta \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}} \\ & =(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})-(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \\ & \overrightarrow{\Delta \mathrm{r}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \therefore \mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\Delta \mathrm{r}} \end{aligned}                                            

\begin{aligned} &\begin{aligned} & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\ & =15-10+35 \\ & =5+35 \end{aligned}\\ &\mathrm{W}=40 \mathrm{~J} \end{aligned}

 

Posted by

sudhir.kumar

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 Mains 2025 Paper 2 Answer Key LIVE: NTA final answer key PDF at jeemain.nta.nic.in soon; BArch result date
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation
anam-khan

Four Kashmiri girls defy the odds, score 99+ percentile in JEE Main 2025

Latest Question