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Question
A force →F=−K(y^i+x^j) (where K is a positive constant) acts on a particle moving in the x - y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is
A force →F=−K(y^i+x^j) (where K is a positive constant) acts on a particle moving in the x - y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is
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Solution
The correct option is A −Ka2
For motion of the particle from (0,0) to (a,0)
Displacement →r=(a^i+0^j)−(0^i+0^j)=a^i
So, work done from (0,0) to (a,0) is given by
W1=→F.→r=−K(y^i+x^j).(a^i)=−Kya
W1=0 (∵y=0 during this motion)
For motion from (a,0) to (a,a) is given by
W2=−K(y^i+x^j).(0^i+a^j)=−Kxa
=−Ka2 (∵x=a during this motion)
∴ Total work done by the forceW=−Ka2+0
W=−Ka2
For motion of the particle from (0,0) to (a,0)
Displacement →r=(a^i+0^j)−(0^i+0^j)=a^i
So, work done from (0,0) to (a,0) is given by
W1=→F.→r=−K(y^i+x^j).(a^i)=−Kya
W1=0 (∵y=0 during this motion)
For motion from (a,0) to (a,a) is given by
W2=−K(y^i+x^j).(0^i+a^j)=−Kxa
=−Ka2 (∵x=a during this motion)
∴ Total work done by the forceW=−Ka2+0
W=−Ka2
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