Question

A force F=-k(^hati + x^hatj) (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 F on the particle is (a) -2ka^(2) , (b) 2ka^(2) , (c)-ka^(2) , (d) ka^(2)

Answer 1

The Total Work Done by the force F on the particle is = -2ka^2

->   Given: Force (F) = -(ki + kxj); Displacement (s) = ai + aj.

->   Work =  ∫F.ds = - ∫(ki + kxj).(ai + aj)ds = - ∫kadx - ∫xkady = -ka^2 - ka^2

                = -2ka^2

Answer 2

Hence the value f net work done is −ka^2

Option (C) is correct.

Explanation:

While moving from (0,0) to (a,0)

• Force is in negative y-direction while displacement is in positive x-direction.  
• W1 = 0
• As force is perpendicular to displacement thus particle moves from (a,0) to (a,a) along a line parallel to y-axis (x=+a)
• The first component of force, −kyiˆ is along negative x-direction (−iˆ) while displacement is in positive y-direction (a,0) to (a,a).
• The second component of force i.e. −kajˆ will perform negative work

W2 = (−kajˆ) (ajˆ) = (−ka) (a) = −ka^2

so net work done on the particle W = W1 + W2

Net work done =0 + (−ka^2) = −ka^2

Hence the value f net work done is −ka^2