Answer
(a) weight = 196 N
$F_N = 196 ~N$
(b) The normal force that the table exerts on the 20-kg box is 294 N.
The normal force that the 20-kg box exerts on the 10-kg box is 98 N.
Work Step by Step
(a) $weight = mg$
$weight = (20.0 ~kg)(9.80 ~m/s^2)$
$weight = 196 ~N$
Since the box is at rest, the normal force which opposes the weight must have the same magnitude as the weight.
$F_N = 196 ~N$
(b) With the 10.0-kg on the 20.0-kg box, the 20.0-kg box pushes down on the table with the weight from a total of 30.0 kg. By Newton's third law, the normal force from the table must be equal and opposite.
The normal force that the table exerts on the 20-kg box:
$F_N = (30.0 ~kg)(9.80 ~m/s^2)$
$F_N = 294 ~N$
The normal force that the 20-kg box exerts on the 10-kg box:
$F_N = (10.0 ~kg)(9.80 ~m/s^2)$
$F_N = 98.0 ~N$
