#### Answer

(a) $W = 206~J$
(b) $P = 68.7~watts$

#### Work Step by Step

(a) Since the block moves at a steady speed, the applied force $F$ must be equal in magnitude to the force of kinetic friction. We can find the magnitude of the force $F$ which is applied by the person. We can use $\mu_k = 0.70$. Therefore;
$F = F_f$
$F = mg~\mu_k$
$F = (10~kg)(9.80~m/s^2)(0.70)$
$F = 68.6~N$
We can find the distance $d$ the block moves in 3.0 seconds.
$d = v~t$
$d = (1.0~m/s)(3.0~s)$
$d = 3.0~m$
We can find the work done by the person.
$W = F~d$
$W = (68.6~N)(3.0~m)$
$W = 206~J$
(b) We can find the power output.
$P = \frac{W}{t}$
$P = \frac{206~J}{3.0~s}$
$P = 68.7~watts$