Question

Find the work done by a force \[\vec F = x\hat i + xy\hat j\] acting a particle to displace it from point \[O\left( {0,0} \right)\] to\[C\left( {2,2} \right)\].

Answer 1

Hint:Work done is obtained by performing the dot product of Force and displacement of the object. Work done is a scalar quantity. The work done is equal to the magnitude of the force that is multiplied by the distance an object moves in the direction of the force.

Formula Used:
\[d\vec w = \vec F.d\vec s\]
Where:
\[d\vec w\]= The amount of work done
\[\vec F\]= The force applied
\[d\vec s\]= The amount of displacement caused due to the applied force.

Complete step by step solution:
Work done can be defined as work is done, when a force acts on an object and it causes a displacement.
In order to calculate the amount of work done, we require three quantities namely, force, displacement and the angle between force and displacement.
In the given question:
Force, \[\vec F = x\hat i + xy\hat j\]
Let us consider the displacement caused by the object be denoted by\[d\vec s\], such that
\[d\vec s = dx\hat i + dy\hat j\]
Let us consider, force \[\vec F\]causes a small displacement of \[d\vec s\]as a result of which\[d\vec w\].
We know:
\[d\vec w = \vec F.d\vec s\]
Putting the values of \[\vec F\]and\[d\vec s\], we find:
\[d\vec w = \left( {x\hat i + xy\hat j} \right).\left( {dx\hat i + dy\hat j} \right)\]
By solving the dot product, we obtain the following equation:
\[d\vec w = x dx + xy dy\]
Now, we need to integrate\[d\vec w\], to get the work done W, thus on the right hand side, we integrate \[d\vec x\]and \[d\vec y\]within their respective limits.
The points \[O\left( {0,0} \right)\]and \[C\left( {2,2} \right)\]are given, integrating \[d\vec x\] from \[0to2\]and integrating \[d\vec y\]from\[0to2\], in order to get the total work done, we obtain:
\[W = _0^2\smallint xdx + _0^2\smallint xydy\]
Thus, we get:
\[W = 6J\]
This is our required answer.

Note:The SI unit for work done is in joules\[(J)\] as work done is a measure of transfer of energy. The integration done above, is done within the two points O and C, owing to their respective coordinate values. Work being a scalar quantity has only magnitude and no direction.