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Question: A force <img src="https://cdn.pureessence.tech/canvas_385.png?top_left_x=1080&top_left_y=0&width=300...

A force (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 forces F⃗\vec { F } on the particle is

A

−2Ka2- 2 K a ^ { 2 }

B
C

−Ka2- K a ^ { 2 }

D

Ka2K a ^ { 2 }

Answer

−Ka2- K a ^ { 2 }

Explanation

Solution

For motion of the particle form (0, 0) to (a, 0)

F⃗=−K(0i^+aj^)\vec { F } = - K ( 0 \hat { i } + a \hat { j } )

Displacement

So work done from (0, 0) to (a, 0) is given by

For motion (a, 0) to (a, a)

and displacement

So work done from (a, 0) to (a, a) W=F⃗⋅r⃗W = \vec { F } \cdot \vec { r }

So total work done =−Ka2= - K a ^ { 2 }