#### Answer

(a) The kinetic energy of the sack is $0.70~J$
(b) The speed of the sack is $0.37~m/s$

#### Work Step by Step

(a) We can find the work that Sam does on the sack:
$Work = F~d$
$Work = (2.0~N)(0.35~m)$
$Work = 0.70~J$
The work that Sam does on the sack is equal to the change in kinetic energy of the sack. Therefore, the kinetic energy of the sack is $0.70~J$
(b) We can find the speed of the sack:
$KE = 0.70~J$
$\frac{1}{2}mv^2 = 0.70~J$
$v = \sqrt{\frac{(2)(0.70~J)}{m}}$
$v = \sqrt{\frac{(2)(0.70~J)}{10.0~kg}}$
$v = 0.37~m/s$
The speed of the sack is $0.37~m/s$