Answer
(a) (i) W = 9.00 J
(ii) W = -9.00 J
(b) (i) Gravity does no work on the 20.0-N block.
(ii) W = 9.00 J
(iii) W = -9.00 J
(iv) The normal force does no work on the 20.0-N block.
(c) The total work done on each block is zero.
Work Step by Step
(a) (i) $W = mg~d = (12.0~N)(0.750~m)$
$W = 9.00~J$
(ii) Since the speed is constant, the tension $T$ is equal to 12.0 N.
$W = (12.0~N)(0.750~m)~cos(180^{\circ}) = -9.00~J$
(b) (i) Since gravity acts at a $90^{\circ}$ to the direction of motion, gravity does no work on the 20.0-N block.
(ii) $W = (12.0~N)(0.750~m) = 9.00~J$
(iii) Since the speed is constant, the force of friction $F_f$ is equal to 12.0 N.
$W = (12.0~N)(0.750~m)~cos(180^{\circ}) = -9.00~J$
(iv) Since the normal force acts at a $90^{\circ}$ to the direction of motion, the normal force does no work on the 20.0-N block.
(c) The total work done on each block is zero.
Note that the kinetic energy of each block does not change.
