Answer
(a) Please see the work below.
(b) $1.11m/s^2$
Work Step by Step
(a) When the force is applied in the forward direction on the tassel then, due to inertia, the string and the tassel want to remain at rest and thus it experiences a backward force. As a result, the tassel moves backward.
(b) We know that
$\Sigma F_y=Tcos\theta -mg =0$
$\implies T=\frac{mg}{cos\theta}$
Similarly $\Sigma F_x=Tsin\theta=ma$
$\implies \Sigma F_x=(\frac{mg}{cos\theta})sin\theta=mgtan\theta=ma$
$\implies a=gtan\theta$
We plug in the known values to obtain:
$a=(9.81)tan6.44^{\circ}$
$a=1.11m/s^2$
