A force F =-k(y hat i +x hat 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

Question

A force F =-k(y hat i +x hat 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) -2 k a2 (B) 2 k a2 (C) -k a2 (D) k a2

Tardigrade Admin · Accepted Answer

For the first displacement y=0 Hence, Fx=0 and no work is done. For the second displacement, Fy=-k a and Δ y=a Work =Fy Δ y=-k a2

$- 2 k a^{2}$

$2 k a^{2}$

$- k a^{2}$

$k a^{2}$

For the first displacement $y = 0$
Hence, $F_{x} = 0$ and no work is done.
For the second displacement, $F_{y} = - ka$ and $Δ y = a$
Work $= F_{y} Δ y = - k a^{2}$

−2ka2

2ka2

−ka2

ka2

For the first displacement y=0 Hence, Fx​=0 and no work is done. For the second displacement, Fy​=−ka and Δy=a Work =Fy​Δy=−ka2

$- 2 k a^{2}$

$2 k a^{2}$

$- k a^{2}$

$k a^{2}$

For the first displacement $y = 0$
Hence, $F_{x} = 0$ and no work is done.
For the second displacement, $F_{y} = - ka$ and $Δ y = a$
Work $= F_{y} Δ y = - k a^{2}$