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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 ______ JJ.

Correct answer

0

Step-by-step solution

Work done by a constant force is given by W=Force(F→)⋅Displacement(r→)W = Force\left(\overset{\to}{F}\right) \cdot Displacement\left(\overset{\to}{r}\right), where, r→=r→2−r→1\overset{\to}{r} = \overset{\to}{r}2 - \overset{\to}{r}1

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, W=F→⋅(r→2−r→1)W = \overset{\to}{F} \cdot \left(\overset{\to}{r}2 - \overset{\to}{r}1\right)

= ( 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]

⇒W=40J⇒ W = 40J

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