Question

How do i prove this?

a particle of mass `m` moves under the influence of the force
`F= a(sinωt i+cosωtj)` if the particle is initially at rest at the origin, prove that the work done on the particle up to time `t` is given by `(a^2 /(mω^2))(1-cosωt)`

Answer 1

`vec F = a{\sin\omegat\hati + \cos\omegat\hatj};`
Newton's Second Law (constant mass):
`vec F = m(dvecv)/(dt);`
`d\vecv = vecF/mdt;`
`\int_{vecv_0}^{vecv(t)}dvecv = a/m{[\int_0^t\sin\omegatdt]\hati + [\int\cos\omegatdt]\hatj}`
`vecv(t)-vecv_0 = a/m{[-(\cos\omegat)/\omega]_0^t \hati + [(\sin\omegat)/(\omega)]_0^t\hatj}`
It is given that the object is initially at rest: `\quad vecv_0 = vec0`

`vecv(t) = a/(m\omega){(1-\cos\omegat)\hati + \sin\omegat\hatj}`

Work Done: `\quad W = \int_{vecr_i}^{vecr_f} vecF.dvecr`
But `\quad vecv = (dvecr)/(dt); \qquad dvecr = vecvdt`

`W = \int_0^tvecF.vecvdt`

`vecF.vecv = a{\sin\omegat\hati + \cos\omegat\hatj}.a/(m\omega){(1-\cos\omegat)\hati+\sin\omegat\hatj}`
`vecF.vecv = a^2/(m\omega){\sin\omegat(1-\cos\omegat) + \cos\omegat\sin\omegat}`
`\qquad \qquad \quad = a^2/(m\omega)\sin\omegat`

`W = a^2/(m\omega)\int_0^t\sin\omegatdt = a^2/(m\omega)[-(\cos\omegat)/\omega]_0^t`
`W = (a^2/(m\omega^2))(1-\cos\omegat)`