#### Answer

a)$W_{f, man} = 2000 \ lbf.ft = 2.57BTU$
b)$W_{f, block} = -2000 \ lbf.ft = -2.57BTU$

#### Work Step by Step

The total work of the man-block system is zero, which means that the work done by the man must be equal to the work done on the block, in magnitude.
I) For the block:
Since the plane is horizontal: $F_{W} = F_{N}$
The equation for the friction work is :
$W_f = \int F_f dx $
$W_f = \mu F_N\Delta x$
With
$\mu = 0.2$
$F_N = 100\ lbf$
$\Delta x = 100\ ft$
$W_f = 2000 \ lbf.ft$
$W_f = 2000 \ lbf.ft \times \frac{1\ BTU}{778.2\ lbf.ft} = 2.57 BTU$
Using the formal sign convention:
$W_{f, block} = -2000 \ lbf.ft = -2.57BTU$
$W_{f, man} = 2000 \ lbf.ft = 2.57BTU$