Answer

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Hint: In this question, we use the formula of work done by the force. when a force is applied to an object, causing displacement. When the force is represented by the vector \[\mathop F\limits^ \to \] and the displacement is represented by the vector \[\mathop S\limits^ \to \] then the work done W is given by the formula \[W = \mathop F\limits^ \to .\mathop S\limits^ \to = \left| F \right|\left| S \right|\cos \theta \].

Complete step-by-step answer:

A particle acted on two constant forces \[\mathop {{F_1}}\limits^ \to = 3i + j - k\] and \[\mathop {{F_2}}\limits^ \to = 4i + j - 3k\] .

So, the net force applied on particle is \[\mathop F\limits^ \to = \mathop {{F_1}}\limits^ \to + \mathop {{F_2}}\limits^ \to \]

$

\Rightarrow \mathop F\limits^ \to = \left( {3i + j - k} \right) + \left( {4i + j - 3k} \right) \\

\Rightarrow \mathop F\limits^ \to = 7i + 2j - 4k \\

$

Now, the particle displaced from the point \[\mathop {{r_1}}\limits^ \to = i + 2j + 3k\] to \[\mathop {{r_2}}\limits^ \to = 5i + 4j + k\]

So, the displacement covered by particle \[\mathop S\limits^ \to = \mathop {{r_2}}\limits^ \to - \mathop {{r_1}}\limits^ \to \]

\[

\Rightarrow \mathop S\limits^ \to = \left( {5i + 4j + k} \right) - \left( {i + 2j + 3k} \right) \\

\Rightarrow \mathop S\limits^ \to = 4i + 2j - 2k \\

\]

Now, we apply formula of work done \[W = \mathop F\limits^ \to .\mathop S\limits^ \to \]

$ \Rightarrow W = \left( {7i + 2j - 4k} \right).\left( {4i + 2j - 2k} \right)$

We know in dot product $i.i = 1,j.j = 1,k.k = 1

i.j=0, i.k=0$

$

\Rightarrow W = 28 + 4 + 8 \\

\Rightarrow W = 40J \\

$

So, the total work done by the forces is 40 Joules.

Note: Whenever we face such types of problems we use some important points. First we find the net force acted on the particle and also the displacement covered by the particle then apply the formula of work done. So, after using dot product we will get the required answer.

Complete step-by-step answer:

A particle acted on two constant forces \[\mathop {{F_1}}\limits^ \to = 3i + j - k\] and \[\mathop {{F_2}}\limits^ \to = 4i + j - 3k\] .

So, the net force applied on particle is \[\mathop F\limits^ \to = \mathop {{F_1}}\limits^ \to + \mathop {{F_2}}\limits^ \to \]

$

\Rightarrow \mathop F\limits^ \to = \left( {3i + j - k} \right) + \left( {4i + j - 3k} \right) \\

\Rightarrow \mathop F\limits^ \to = 7i + 2j - 4k \\

$

Now, the particle displaced from the point \[\mathop {{r_1}}\limits^ \to = i + 2j + 3k\] to \[\mathop {{r_2}}\limits^ \to = 5i + 4j + k\]

So, the displacement covered by particle \[\mathop S\limits^ \to = \mathop {{r_2}}\limits^ \to - \mathop {{r_1}}\limits^ \to \]

\[

\Rightarrow \mathop S\limits^ \to = \left( {5i + 4j + k} \right) - \left( {i + 2j + 3k} \right) \\

\Rightarrow \mathop S\limits^ \to = 4i + 2j - 2k \\

\]

Now, we apply formula of work done \[W = \mathop F\limits^ \to .\mathop S\limits^ \to \]

$ \Rightarrow W = \left( {7i + 2j - 4k} \right).\left( {4i + 2j - 2k} \right)$

We know in dot product $i.i = 1,j.j = 1,k.k = 1

i.j=0, i.k=0$

$

\Rightarrow W = 28 + 4 + 8 \\

\Rightarrow W = 40J \\

$

So, the total work done by the forces is 40 Joules.

Note: Whenever we face such types of problems we use some important points. First we find the net force acted on the particle and also the displacement covered by the particle then apply the formula of work done. So, after using dot product we will get the required answer.

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