What Happens to the Normal Force When the Applied Force is Removed?

[Sinusoidal]

Homework Statement

A force of 10.0 N is applied at an angle of 30° to the horizontal on a 1.25-kg block initially at rest on a frictionless surface. a) What is the magnitude of the block’s acceleration? b) What is the magnitude of the normal force?

This was en example in my book. I understood this part. The answer to a) is 6.93 m/s^{2} , and the answer to b) is 17.3 N.

I didn’t understand the follow-up exercise.

a) Suppose the applied force on the block is applied for only a short time. What is the magnitude of the normal force after the applied force is removed? b) If the block slides off the edge of the table, what would be the net force on the block just after it leaves the table (with the applied force removed)?

The book gives these answers:

a) 7.35 N
b) Neglecting air resistance, 7.35 N, downward

Homework Equations

I thought you'd use this since the applied force was removed, but I'm not getting the right answer.

#1 ∑Fy=N-mg=ma=0
N=mg


#2 ∑Fy=N-Fy-mg=ma=0

The Attempt at a Solution

N=(1.25kg)(9.80 m/s^{2})=12.3 N

Earlier, I used the 2nd equation when I was calculating the normal force with the applied force, so I thought removing Fy from the equation would give me the correct answer. Help!

Thanks!

 

[PhanthomJay]

Looks like you're right and the book answer is incorrect.