An object is displaced from position vector ${{r}_{1}}=(2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,)m$ to \[{{r}_{2}}=(4\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,)m\] under a force$F=(3{{x}^{2}}\overset{\hat{\ }}{\mathop i}\,+2y\overset{\hat{\ }}{\mathop{j}}\,)$. Find the work done by this force.
Answer
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Hint: Use formula of work for infinitesimal displacement. Use integration to solve equations. Compare the above example to daily life practical examples to make it easy to understand.
Complete Step-by-Step solution:
First understand what this question wants to say. Then predict what should be our answer.
According to the question, one object which is displaced from its initial position to the next position by some force F.
Initial position is given by,
${{r}_{1}}=(2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,)m$
Next position is given by,
\[{{r}_{2}}=(4\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,)m\]
Force given is,
$F=(3{{x}^{2}}\overset{\hat{\ }}{\mathop i}\,+2y\overset{\hat{\ }}{\mathop{j}}\,)$
Work done by an object when it moves from one place to another is given by
\[\begin{align}
& W=\int Fds \\
& where,ds=\overset{\hat{\ }}{\mathop{i}}\,dx+\overset{\hat{\ }}{\mathop{j}}\,dy \\
& therefore, \\
& W=\int {}_{2}^{0}3{{x}^{2}}dx+\int {}_{3}^{4}3y\partial y \\
\end{align}\]
(Since i vector moves from 2 to 0 and j vector moves 3 to 4)
\[\begin{align}
& W=[{{x}^{3}}]_{2}^{0}+[{{y}^{2}}]_{3}^{4} \\
& by solving \\
& W=-8+16-9 \\
& W=-1joule \\
\end{align}\]
Work done by an object from one place to another by a force is -1 joule.
Additional information:
Work is the product of force and displacement. Work is said to be positive when force has a component in the direction of displacement. International unit of system i.e. S.I unit of work is joule (J).
Note: Unit of work is joule. There are lots of conversions of joule. i and j are vectors.
Dot product of vector
$\begin{align}
& \overset{\hat{\ }}{\mathop{i}}\,.\overset{\hat{\ }}{\mathop{i}}\,=1 \\
& \overset{\hat{\ }}{\mathop{j}}\,.\overset{\hat{\ }}{\mathop{j}}\,=1 \\
& \overset{\hat{\ }}{\mathop{k}}\,.\overset{\hat{\ }}{\mathop{k}}\,=1 \\
\end{align}$
Cross product of vector
\[\begin{align}
& \overset{\hat{\ }}{\mathop{i}}\,\times \overset{\hat{\ }}{\mathop{i}}\,=0 \\
& \overset{\hat{\ }}{\mathop{j}}\,\times \overset{\hat{\ }}{\mathop{j}}\,=0 \\
& \overset{\hat{\ }}{\mathop{k}}\,\times \overset{\hat{\ }}{\mathop{k}}\,=0 \\
\end{align}\]
Complete Step-by-Step solution:
First understand what this question wants to say. Then predict what should be our answer.
According to the question, one object which is displaced from its initial position to the next position by some force F.
Initial position is given by,
${{r}_{1}}=(2\overset{\hat{\ }}{\mathop{i}}\,+3\overset{\hat{\ }}{\mathop{j}}\,)m$
Next position is given by,
\[{{r}_{2}}=(4\overset{\hat{\ }}{\mathop{j}}\,+6\overset{\hat{\ }}{\mathop{k}}\,)m\]
Force given is,
$F=(3{{x}^{2}}\overset{\hat{\ }}{\mathop i}\,+2y\overset{\hat{\ }}{\mathop{j}}\,)$
Work done by an object when it moves from one place to another is given by
\[\begin{align}
& W=\int Fds \\
& where,ds=\overset{\hat{\ }}{\mathop{i}}\,dx+\overset{\hat{\ }}{\mathop{j}}\,dy \\
& therefore, \\
& W=\int {}_{2}^{0}3{{x}^{2}}dx+\int {}_{3}^{4}3y\partial y \\
\end{align}\]
(Since i vector moves from 2 to 0 and j vector moves 3 to 4)
\[\begin{align}
& W=[{{x}^{3}}]_{2}^{0}+[{{y}^{2}}]_{3}^{4} \\
& by solving \\
& W=-8+16-9 \\
& W=-1joule \\
\end{align}\]
Work done by an object from one place to another by a force is -1 joule.
Additional information:
Work is the product of force and displacement. Work is said to be positive when force has a component in the direction of displacement. International unit of system i.e. S.I unit of work is joule (J).
Note: Unit of work is joule. There are lots of conversions of joule. i and j are vectors.
Dot product of vector
$\begin{align}
& \overset{\hat{\ }}{\mathop{i}}\,.\overset{\hat{\ }}{\mathop{i}}\,=1 \\
& \overset{\hat{\ }}{\mathop{j}}\,.\overset{\hat{\ }}{\mathop{j}}\,=1 \\
& \overset{\hat{\ }}{\mathop{k}}\,.\overset{\hat{\ }}{\mathop{k}}\,=1 \\
\end{align}$
Cross product of vector
\[\begin{align}
& \overset{\hat{\ }}{\mathop{i}}\,\times \overset{\hat{\ }}{\mathop{i}}\,=0 \\
& \overset{\hat{\ }}{\mathop{j}}\,\times \overset{\hat{\ }}{\mathop{j}}\,=0 \\
& \overset{\hat{\ }}{\mathop{k}}\,\times \overset{\hat{\ }}{\mathop{k}}\,=0 \\
\end{align}\]
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