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Find the work W done by the force F in moving an object from a point A in space to a point B in space is defined as W = F.. Find the work done by a force of 3 newtons acting in the direction 2i + j +2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).

The aim of this question is to develop a concrete understanding of the key concepts related to vector algebra such as magnitude, direction, and the dot product of two vectors in cartesian form.

Given a vector $ \vec{ A } \ = \ a_1 \hat{ i } \ + \ a_2 \hat{ j } \ + \ a_3 \hat{ k } $, its direction and magnitude are defined by the following formulas:

\[ |A| \ = \ \sqrt{ a_1^2 \ + \ a_2^2 \ + \ a_3^2 } \]

\[ \hat{ A } \ = \ \dfrac{ \vec{ A } }{ |A| } \]

The dot product of two vectors $ \vec{ A } \ = \ a_1 \hat{ i } \ + \ a_2 \hat{ j } \ + \ a_3 \hat{ k } $ and $ \vec{ B } \ = \ b_1 \hat{ i } \ + \ b_2 \hat{ j } \ + \ b_3 \hat{ k } $ is defined as:

\[ \vec{ A }.\vec{ B } \ = \ a_1 b_1 \ + \ a_2 b_2 \ + \ a_3 b_3 \]

Expert Answer

Let:

\[ \vec{ A } \ = \ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \]

To find the direction of $ \vec{ A } $, we can use the following formula:

\[ \text{ Direction of } \vec{ A } = \ \hat{ A } \ = \ \dfrac{ \vec{ A } }{ |A| } \]

\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ ( 2 )^2 \ + \ ( 1 )^2 \ + \ ( 2 )^2 } } \]

\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ 4 \ + \ 1 \ + \ 4 } } \]

\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ \sqrt{ 9 } } \]

\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } }{ 3 } \]

\[ \Rightarrow \hat{ A } \ = \ \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \]

Given that:

\[ \text{ Magnitude of Force } = \ |F| = 3 \ N \]

\[ \text{ Direction of Force } = \ \hat{ F } \ = \ \hat{ A } \ = \ \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \]

To find $ \vec{ F } $ we can use following formula:

\[ \vec{ F } \ = \ |F| . \hat{ F } \]

\[ \Rightarrow \vec{ F } \ = \ ( 3 ) . \bigg ( \dfrac{ 2 }{ 3 } \hat{ i } \ + \ \dfrac{ 1 }{ 3 } \hat{ j } \ + \ \dfrac{ 2 }{ 3 } \hat{ k } \bigg ) \]

\[ \Rightarrow \vec{ F } \ = \ 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \]

To find $ \vec{ AB } $ we can use following formula:

\[ \Rightarrow \vec{ AB } \ = \ \bigg ( 0 \hat{ i } \ + \ 2 \hat{ j } \ + \ 0 \hat{ k } \bigg ) \ – \ \bigg ( 0 \hat{ i } \ + \ 0 \hat{ j } \ + \ 0 \hat{ k } \bigg ) \]

\[ \Rightarrow \vec{ AB } \ = \ 2 \hat{ j } \]

To find the work done $ W $, we can use the following formula:

\[ W \ = \ \vec{ F } . vec{ AB } \]

\[ \Rightarrow W \ = \ \bigg ( 2 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } \bigg ) .  \bigg ( 2 \hat{ j } \bigg ) \]

\[ \Rightarrow W \ = \ ( 2 )( 0 ) \ + \ ( 1 )( 2 ) \ + \ ( 2 )( 0 ) \]

\[ \Rightarrow W \ = \ 2 \ J \]

Numerical Result

\[ W \ = \ 2 \ J \]

Example

Given $ \vec{ F } \ = \ 2 \hat{ i } \ + \ 4 \hat{ j } \ + \ 2 \hat{ k } $ and $ \vec{ AB } \ = \ 7 \hat{ i } \ + \ \hat{ j } \ + \ 2 \hat{ k } $, Find the work done $ \vec{ W }.

To find $ W $, we can use following formula:

\[ W \ = \ \vec{ F } . vec{ AB } \]

\[ \Rightarrow W \ = \ \bigg ( 2 \hat{ i } \ + \ 4 \hat{ j } \ + \ 2 \hat{ k } \bigg ) .  \bigg ( 7 \hat{ i } \ + \ 1 \hat{ j } \ + \ 2 \hat{ k } \bigg )\]

\[ \Rightarrow W \ = \ ( 2 )( 7 ) \ + \ ( 4 )( 1 ) \ + \ ( 2 )( 2 ) \]

\[ \Rightarrow W \ = \ 22 \ J \]

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