Monday, July 2, 2007

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.

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