Answer
(a) $42.2^{\circ}$
(b) $14.8Kg$
Work Step by Step
(a) We can find the required angle as
$W=Fdcos\theta$
This can be rearranged as:
$\theta=cos^{-1}(\frac{W}{Fd})$
We plug in the known values to obtain:
$\theta=cos^{-1}(\frac{50.0}{45.0(1.50)})$
$\theta=42.2^{\circ}$
(b) The mass of the block can be determined as:
$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$W=\frac{1}{2}mv_f^2-\frac{1}{2}m(0)^2$
This can be rearranged as:
$m=\frac{2W}{v_f^2}$
We plug in the known values to obtain:
$m=\frac{2(50.0)}{2.60}$
$m=14.8Kg$
