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Question: A force \(F = - \left( {yi\, + xj} \right)\) acts on a particle moving in the X-Y plane. Starting fr...

A force F=−(yi +xj)F = - \left( {yi\, + xj} \right) acts on a particle moving in the X-Y plane. Starting from the origin, the particle is taken along the position X-axis to the point (2a,0) and then parallel to the Y-axis to the point (2a,2a). The total work done by the particle is
A. −4a2- 4{a^2}
B. −2a2- 2{a^2}
C. 4a24{a^2}
D. 2a22{a^2}

Explanation

Solution

Applying work done and force relation formula we can calculate the solution of the given problem. First calculate the work done when the particles start moving from origin along the X-axis then calculate the work done by the p[aricles when they move parallel to the y-axis. Then adding both the work done we will get the total work done by the particle.

Complete step by step answer:
As per the problem a force acts on a particle moving in the X-Y plane. The particle starts moving from the origin, the particle is taken along the position X-axis to the point (2a,0) and then the particle moves parallel to the Y-axis to the point (2a,2a).We need to calculate the total work done by the particle due to the applied force.From this above statement we can conclude that two different works are done by the particles first along the X- axis and the second work is done due to the Y-axis.

Hence, total work done=work done by the particles due to X-axis + work done by the particles due to Y-axis
W=W1+W2W = {W_1} + {W_2}
We know,
W=∫F⋅drW = \int {F \cdot dr}
Where, Work done=WW, Force applied on the particle=FF and change in position due to applied force=drdr
First calculate work done by the particles when the particles move along the position X-axis to the point (2a,0) we get,
W1=∫02aF⋅dr{W_1} = \int_0^{2a} {F \cdot dr}
Putting the value of F from the problem,
⇒W1=∫02a−(yi +xj)⋅(dxi +dyj)\Rightarrow {W_1} = \int_0^{2a} { - \left( {yi\, + xj} \right)} \cdot \left( {dxi\, + dyj} \right)
Here y=0 as the particle is moving only along X-axis,
Hence dy=0,
W1=∫02a−(yi )⋅(dxi ){W_1} = \int_0^{2a} { - \left( {yi\,} \right)} \cdot \left( {dxi\,} \right)
As y=0 we get,
W1=0……(1){W_1} = 0 \ldots \ldots \left( 1 \right)
Now calculating the work done when the particle is parallel to the Y-axis to the point (2a,2a) we get,
W2=∫02aF⋅dr{W_2} = \int_0^{2a} {F \cdot dr}

Initially the particle is at the position (2a,0) and now it moves to (2a,2a) hence the change in Y-axis is from 0 to 2a therefore the limite of the integration is from 0 to 2a
W2=∫02a−(yi +xj)⋅(dxi +dyj){W_2} = \int_0^{2a} { - \left( {yi\, + xj} \right)} \cdot \left( {dxi\, + dyj} \right)
⇒W2=∫02a−(ydx+xdy)\Rightarrow {W_2} = \int_0^{2a} - \left( {ydx + xdy} \right)
Here x=2a then dx=0 as x is a constant value,we get
W2=∫02a−(y(0)+2ady){W_2} = \int_0^{2a} - \left( {y\left( 0 \right) + 2ady} \right)
⇒W2=∫02a−(2ady)\Rightarrow {W_2} = \int_0^{2a} - \left( {2ady} \right)
⇒W2=−2a∫02ady\Rightarrow {W_2} = - 2a\int_0^{2a} {dy}
⇒W2=−2a[y]02a\Rightarrow {W_2} = - 2a\left[ y \right]_0^{2a}
⇒W2=−2a[2a−0]\Rightarrow {W_2} = - 2a\left[ {2a - 0} \right]
⇒W2=−4a2……(2)\Rightarrow {W_2} = - 4{a^2} \ldots \ldots \left( 2 \right)
Hence total work done will be,
W=W1+W2W = {W_1} + {W_2}
Putting equation (1)\left( 1 \right) and (2)\left( 2 \right)we get,
W=0+(−4a2)W = 0 + \left( { - 4{a^2}} \right)
∴W=−4a2\therefore W = - 4{a^2}

Therefore the correct option is (A)\left( A \right).

Note: Whenever a particle moves from one place to another with the help of some external force then some amount of work is done on the body. Remember the total work done by the particle is the sum of all the work done by the particles at different positions of the coordinate axis.