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.

 

[baba89]

Hello,
Thank you for your question. It seems like you are on the right track with your attempt at a solution. Let me guide you through the steps to find the correct answers for both parts a) and b).
a) When the applied force is removed, the block will continue to move with a constant velocity, as there are no other forces acting on it. This means that the net force on the block is equal to zero. We can use this information to set up the following equation:
∑Fy = N - mg = ma = 0
Where N is the normal force, m is the mass of the block, g is the acceleration due to gravity, and a is the acceleration of the block.
We can rearrange this equation to solve for N:
N = mg
Substituting in the values given in the problem, we get:
N = (1.25 kg)(9.80 m/s^2) = 12.25 N
This is the normal force when the applied force is still present. To find the normal force after the applied force is removed, we need to use the same equation, but with a different value for the acceleration. Since the block is now moving with a constant velocity, its acceleration is equal to zero.
N - mg = ma = 0
N = mg
N = (1.25 kg)(9.80 m/s^2) = 12.25 N
Therefore, the magnitude of the normal force after the applied force is removed is also 12.25 N.

b) When the block slides off the edge of the table, it is no longer in contact with the surface and therefore the normal force is equal to zero. This means that the only force acting on the block is its weight, which is equal to mg.
∑Fy = N - mg = ma = 0
N = mg
N = (1.25 kg)(9.80 m/s^2) = 12.25 N
Therefore, the net force on the block just after it leaves the table is equal to its weight, which is 12.25 N, acting downward.

I hope this helps clarify things for you. Remember, when the applied force is removed, the block will continue to move at a constant velocity, so its acceleration is equal to zero. And when the block is no longer in contact with