Accelerating the Box

Example :

What happens when a person pulls upward on the box with a force equal to, or greater than , the box's weight , say F (p) = 100.0 N rather than the 40.0 N shown in figure.

Solution: The net force on the box is

F (y) = F (N) - mg + F (p)

= F (N) - 98.0 N + 100.0 N

The box accelerates upward, so the net force is not zero.

Normal force is zero.

The net force = F (y) = F (p) - mg = 100.0 N - 98.0 N

= 2.0 N

a (y) = F (y)/ m = 2.0 N/10.0 kg = 0.20 m/s^2.