Question

Determine the work done by a force F = xyi + yzj + xzk in taking a particle along

the path defined by the equation r(t) = ti + 2t^2j + t^3k, 0<= t <= 1 from t=0 to t= 2.
is the force conservative??​

Answer 1

We have to determine the work done by a force F = xy i + yz j + zx k in taking a particle along the path defined by the equation r(t) = t i + 2t² j + t³ k, 0 ≤ t ≤ 1.

solution : first find infinitesimal value of r(t), differentiating r(t) with respect to t,

dr(t) = dti + 4t dt j + 3t² dt k

now work done is the W = ∫F.dr(t)

= ∫(xy i + yz j + zx k).( dti + 2t dt j + 3t² dtk)

= ∫xy dt + ∫ yz 2t dt + ∫zx 3t² dt

we know, r(t) = t i + t² j + t³ k = x i + y j + z k

x = t , y = t² and z = t³

= ∫(t × t²) dt + ∫(t² × t³) 2t dt + ∫(t³ × t) 3t² dt

= [t⁴/4] + 2[t^7/7] + 3[t^7/7]

= [t⁴/4] + 5[t^7/7] ...(1)

now put the limits 0 to 1

= 1/4 + 5/7

= 27/28

Therefore the work done by the force is 27/28 J.

The force will be conservative in a duration of time if work done by force in that duration of time is zero.

so the Workdone by force from 0 to 2s = [t⁴/4] j 5[t^7/7] [ from equation (1). ]

taking limits 0 to 2.

= (2)⁴/4 + 5(2)^7/7

= 4 + 320/7

= 348/7 J ≠ 0

Therefore the force is not conservative.