JEE Main 20231 FebruaryMorning Shift
- Exam
- JEE Main
- Year
- 2023
- Date
- 1 February
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Work, Power and Energy
JEE Main 2023 · 1 February · Morning Shift · Physics · Work, Power and Energy
A small particle moves to position 5 ^ i - 2 ^ j + ^ k from its...
A small particle moves to position 5\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k} from its initial position 2\overset{^}{i} + 3\overset{^}{j} - 4\overset{^}{k} under the action of force 5\overset{^}{i} + 2\overset{^}{j} + 7\overset{^}{k}N. The value of work done will be ______ .
Correct answer
0
Step-by-step solution
Work done by a constant force is given by , where,
Given here, \overset{\to}{{r}_{1}} = 5\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k} and \overset{\to}{r}2 = 2\overset{^}{i} + 3\overset{^}{j} - 4\overset{^}{k}
So,
= ( 5\overset{^}{i} + 2\overset{^}{j} + 7\overset{^}{k} ) \cdot \left[( 5\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k} ) - ( 2\overset{^}{i} + 3\overset{^}{j} - 4\overset{^}{k} )\right]