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)

1 Answer
Dec 1, 2017

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)