Question

a 1 kg ball is thrown straight up in the air. What is the net force acting on the ball when it reaches its maximum height?

Answer 1

From what I understand, when it reaches its maximum height it stops moving, because that's the furthest up it can go. So there will be no more kinetic energy acting upwards. What's left is the potential energy acting downward.

Answer 2

At any moment? What about when it has velocity?

Answer 3

Gravity is the force acting downwards. The net force will also result downwards, yes, that is why it stops and falls back down. But the magnitude is not just the magnitude of the force downward. It will be the total of mg and ma(upwards).

Answer 4

I guess when talking about magnitude I should say the difference instead of total.

Answer 5

But a is the result of v, as in when there is movement. So whenever the ball is moving, there will always be a except for gravity. It exists even when you are standing still. g is basically the natural a produces by a massive body.

Answer 6

What you showed is the acceleration and that is true. The acceleration will always be g. But what you claimed earlier is that the only force acting on an object thrown into the air is only the its weight. That is not true for when there is velocity (kinetic energy) also acting on the object.

Answer 7

Are you talking about at the maximum height? Then yes there is no v left so no other energy/force acting on it. But you claim that at any point there is only gravity acting on the ball. That is what I am arguing.

Answer 8

Well in any case, @Lafitness3618 , as I was saying since the ball stops moving at that point, it means that both the force acting upwards and downwards are equal. Thus, the net force would be zero. But as Carlos and I were discussing here, the acceleration might be a different case. This might provide you with some understanding: http://answers.google.com/answers/threadview/id/576366.html

Answer 9

Alright but let me ask you this: if the net force is ALWAYS the gravity, acting downwards, then why did the ball even move up in the first place?

Answer 10

But you just said that the net force is downward, meaning that the gravity overpowers the force applied upwards. When net force is downwards, the ball will move down, not up as in this case. If being exerted gravity yet the ball still moves up, isn't it only logical that the force upwards overpowers the gravity? If what you

Answer 11

*if what you say is true, then the ball would immediately fall down, never going up even a little.

Answer 12

Now for gravity though, it is always the acceleration of a free falling object. But we are talking about net force. Net, meaning resulting. Meaning is you say net force is downward, then the ball is moving downwards.

Answer 13

The answer is 9.8 Newtons down towards the center of the Earth. Velocity is zero so there is no frictional force acting on the body leaving only the force of gravity acting on it which is F=MA=(1kg)(9.8m/s^2)=9.8 Newtons

Answer 14

@plainntall if I understand correctly, that is obtained by adding the downward force(including fiction if any) to the upward force, right? Here is my problem with the question. What I had in mind is that exactly at the maximum point, the 2 forces balance each other. There is still mg acting on the ball, but it is equal to the F upwards. Isn't it immediately when the ball starts moving again that the mg overpowers the F upwards again, making the net force mg? I am saying at the top the net force is 0.

Answer 15

There is no force acting up to balance the force of gravity. That is way the ball will begin accelerating downward after it reaches maximum height. If there were a force equal to the force of gravity the ball would stay at the maximum height since the net force is zero and not 9.8 newtons downward.

The force that was acting up was much greater than the force of gravity and resulted in a velocity upward, but that force was immediately removed when the ball was released by the thrower. After that only the frictional force and gravitational force acted upon the ball trying to slow it down. Frictional forces depend on the velocity of the ball, when the velocity is zero the force is zero. When the ball was moving up the frictional force and the gravitational force acted in the same direction. When the ball goes down the frictional force will act upwards while the gravitational force will continue acting downwards.

Answer 16

But if you picture the whole process as graphs between F upwards and mg, shouldn't there be a point where they intersect each other? Because F will go down until it is the same as mg, am I right? That is the point when the ball stops moving. Then F will continue to go down below mg, making the net force side to mg even more. But at that intersection, they are both equal.

Answer 17

If F is the total force then after the ball is thrown the only forces acting on the ball are
F=Fg+Ff where Fg is the force due to gravity and Ff is the force due to friction. The only time Fg and Ff will be equal, but oppositely directed is when the ball has reached terminal velocity while falling down. At that time the net force will indeed be zero, but the velocity will me non-zero.

Please remember that Force is equal to acceleration times mass or F=ma. Acceleration is the change in velocity over time. Velocity can be zero at one moment in time but the change of velocity may not be zero. If the velocity changes from 60m/s to 0m/s in 5 seconds then even though the speed is zero at one moment in time, the acceleration would be 60/5 or 12m/s^2.

Answer 18

Oh firstly I denoted F as just force. Total force is net force for me. What I have been trying to say is that at the point v=0 before the ball falls, the force the ball had(or what's left of it), given by the hand, is equal to Fg(put air resistance aside first). This question is asking about at that point only.

What you said in your second paragraph explains the motion downwards AFTER the maximum height was reached. Your 3rd paragraph gives an example of the acceleration BEFORE the maximum height before it stops. I agree, acceleration exists but during motion where v=0 can only be a starting or ending point to calculating acceleration(i.e we cannot calculate acceleration for a period where an object stops and then moves again, unless we're finding average acceleration).

What I am saying is we need to focus on the question here. It asks for the net force when the ball stops moving at the maximum height.

Answer 19

Oh my goodness no wonder you've been going that direction. The ball was not released from top. It was thrown from bottom, traveled to the top, then goes back down.

Answer 20

"a 1 kg ball is thrown straight up in the air"

Answer 21

Then what does this mean??? "And we have answered that from the moment the ball is released until it touches the ground, during all that time, the net force acting on it is gravitational force with value: M·g. "

Answer 22

Carlos, I did NOT say there is NO force acting on a stationary object. Please, read my response to plainn. I said that when the object becomes stationary at the top, the two forces are of equal magnitude.

Answer 23

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