# Work done by a variable force

The equation W = F->.S-> = FS cos θ is applicable when remains constant but when the force is variable work is obtained by integrating F->.ds->

Thus, W = ∫F->.ds->

An example of a variable force is the spring force in which force depends on the extension x,

i.e., F->= F->(x)

When the force is time dependent, we have

W = ∫F->.dX-> = ∫F->.v->dt

where F-> and v-> are the instantaneous force and velocity vectors.

Example : A body acted upon by a force F-> , given by, F-> = – k [(cos ωt) i^ + (sin ωt )j^ ] undergoes displacement, where the position vector r-> of the body is given by r-> = a[cos (ωt + α) +sin (ωt + α) ]. Find the work done by the force from time t = 0 to time t = 2π /ω

Solution : The position of body is given by

r-> = a[cos (ωt + α)i^ + sin (ωt + α)j^ ]

Its velocity is given by,
v-> = dr->/dt = d/dt [ a cos (ωt + α) + a sin (ωt + α) ]

= – aω sin (ωt + α)i^ + a ωcos (ωt + α)j^

The power developed by this force is, = (aωk) cos (ωt) sin (ωt + α) – (aωk) sin (ωt) cos (ωt + α)

= aωk [ sin (ωt + α) cos ωt – cos (ωt + α) sin ωt]

= aαk sin (ωt + α – ωt)

= aαk sin α

W = ∫ dW = aωk sin α o2π/ωdt

= aωk sin α x 2π/ω = 2π a k sin α

Exercise 3: A particle is acted upon by a force given by = A cos ωt +B and its position vector is given by r-> = a[(cos (ωt)i^ + sin (ωt)j^ ] + (1/2) bt2k^
Find the work done on the particle by the force F-> from time t = π/ 2ω to time t = π /ω