First slide

Work

Question

A force $F = - K (yi + xj)$ (where K is a positive constant) acts on a particle moving in the xy-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 F on the particles is

Moderate

A

$- 2 {Ka}^{2}$
B

$2 {Ka}^{2}$
C

$- {Ka}^{2}$
D

${Ka}^{2}$

Solution

While moving from (0,0) to (a,0)
Along positive x-axis, y = 0 $∴ \vec{F} = - kx \hat{j}$
i.e. force is in negative y-direction while displacement is in positive x-direction.
$∴ W_{1} = 0$
Because force is perpendicular to displacement
Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis $(x = + a)$ during this $\vec{F} = - k (y \hat{i} + a \hat{J})$
The first component of force, $- ky \hat{i}$ will not contribute any work because this component is along negative x-direction $(- \hat{i})$ while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. $- ka \hat{j}$ will perform negative work
$∴ W_{2} = (- ka \hat{j}) (a \hat{j}) = (- ka) (a) = - {ka}^{2}$
So net work done on the particle $W = W_{1} + W_{2}$
$= 0 + (- {ka}^{2}) = - {ka}^{2}$

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