A 2 kg ball has been thrown vertically upward. If we ignore the air resistance what are the directions and size of the net force acting on the ball while it is traveling upward.
Direction= right, left, up, down?
Size=?
Obviously there are two forces acting on a ball: weight (the force of gravity), which points down, and the throwing force, which points up. If only weight acted on the ball, then it would move down at all times; however, initially it flies up.
It is known since Aristotle that "in the absence of an external motive power, all objects (on Earth) would come to rest and that moving objects only continue to move so long as there is a power inducing them to do so" (Wikipedia). Everybody who has ever pushed a heavy cupboard knows that if there is no force applied, there is no velocity. Even my cat Meowton discovered that as soon as he stops pushing a ball across the carpet, it stops, and in order for the ball to move with constant speed, it must be acted upon by a paw. Moreover, the greater the mass of the ball, the greater the force that has to be applied to generate the same velocity. I call this the first and second laws of Meowton:
(1) If the net force acting on an object is zero, then the velocity of the object is zero.
(2) The velocity of an object is proportional to the net force applied to it and inversely proportional to the mass of the object. If the velocity is denoted by v, net force by F and mass by m, then v = F / m, or F = vm.
In the case of a thrown ball, it follows from the second law of Meowton that weight, which points down, can only generate downward velocity, so we have to postulate throwing force, which acts upward. Let us denote weight by W and throwing force by T. Then the net force is T - W, so (2) implies v = (T - W) / m. Obviously, W and m don't change with time, but v decreases. Therefore, T must decrease with time. Let's assume that
$\displaystyle T(t)=\begin{cases}T_0-ct&\text{if }T_0-ct\ge0\\0&\text{otherwise}\end{cases}$
That is, T decreases by some constant c every second until it is completely used up and becomes zero. Obviously, throwing force, which acts upward, cannot become negative and start acting downward. Moreover, we know that the greater the mass is, the faster the speed decreases. Let's say that c = mg for some coefficient g. Then $\displaystyle T_0-mgt\ge0$ iff $\displaystyle t\le T_0/(mg)$. Altogether,
$\displaystyle T(t)=\begin{cases}T_0-mgt& \text{if }t\le T_0/(mg)\\ 0&\text{otherwise}\end{cases}$
and
v(t) = (T(t) - W) / m.
After throwing force is exhausted and T(t) = 0, the velocity stabilizes at -W / m (the initial minus sign signifies that it points down). That's why it is called terminal velocity.
Let's see how this plays out in an example. Let m = 2, W = 20, T₀ = 30 and g = 10. Then
$\displaystyle v(t)=\begin{cases}(30-20t-20)/2=5-10t&\text{if }t\le30/20=3/2\\-10&\text{if }t>3/2\end{cases}$
The graph of v(t) is the following.
Initially the velocity is 5m/s pointing up. At t = 0.5 the ball reaches the top of its trajectory and starts going down. At t = 1 it passes the initial point and its velocity equals the initial velocity but points in the opposite direction.
There is one open question about the throwing force, though. What causes it to decrease? There are plenty examples of forces that decrease with distance. For example, the closer two magnets are, the stronger they attract or repel each other. Gravity and electric force between two charges also decrease with distance. But a ball that is thrown directly up initially moves away, but then moves closer and passes through the point where it started its movement. Our formula shows that throwing force decreases with time, not distance. Well, I heard that in Einstein's theory of relativity time and distance are parts of the same space–time continuum. But relativistic effects only becomes noticeable when the velocity is close to the speed of light...
To be honest, I have no idea what I am talking about.
Yes, as long as the ball is still in the person's hand, there is an upward force that accelerates it from 0 to its initial velocity. Once it leaves the person's hand, the is only the downward force of gravity.
This is not true. If the net force on an object is zero, then the acceleration is zero.It is known since Aristotle that "in the absence of an external motive power, all objects (on Earth) would come to rest and that moving objects only continue to move so long as there is a power inducing them to do so" (Wikipedia). Everybody who has ever pushed a heavy cupboard knows that if there is no force applied, there is no velocity. Even my cat Meowton discovered that as soon as he stops pushing a ball across the carpet, it stops, and in order for the ball to move with constant speed, it must be acted upon by a paw. Moreover, the greater the mass of the ball, the greater the force that has to be applied to generate the same velocity. I call this the first and second laws of Meowton:
(1) If the net force acting on an object is zero, then the velocity of the object is zero.
Acceleration, not "velocity"!(2) The velocity of an object is proportional to the net force applied to it and inversely proportional to the mass of the object. If the velocity is denoted by v, net force by F and mass by m, then v = F / m, or F = vm.
Was this all a joke that I am too dense to get? I guess I should have tumbled when you talked of "Meowton".In the case of a thrown ball, it follows from the second law of Meowton that weight, which points down, can only generate downward velocity, so we have to postulate throwing force, which acts upward. Let us denote weight by W and throwing force by T. Then the net force is T - W, so (2) implies v = (T - W) / m. Obviously, W and m don't change with time, but v decreases. Therefore, T must decrease with time. Let's assume that
$\displaystyle T(t)=\begin{cases}T_0-ct&\text{if }T_0-ct\ge0\\0&\text{otherwise}\end{cases}$
That is, T decreases by some constant c every second until it is completely used up and becomes zero. Obviously, throwing force, which acts upward, cannot become negative and start acting downward. Moreover, we know that the greater the mass is, the faster the speed decreases. Let's say that c = mg for some coefficient g. Then $\displaystyle T_0-mgt\ge0$ iff $\displaystyle t\le T_0/(mg)$. Altogether,
$\displaystyle T(t)=\begin{cases}T_0-mgt& \text{if }t\le T_0/(mg)\\ 0&\text{otherwise}\end{cases}$
and
v(t) = (T(t) - W) / m.
After throwing force is exhausted and T(t) = 0, the velocity stabilizes at -W / m (the initial minus sign signifies that it points down). That's why it is called terminal velocity.
Let's see how this plays out in an example. Let m = 2, W = 20, T₀ = 30 and g = 10. Then
$\displaystyle v(t)=\begin{cases}(30-20t-20)/2=5-10t&\text{if }t\le30/20=3/2\\-10&\text{if }t>3/2\end{cases}$
The graph of v(t) is the following.
Initially the velocity is 5m/s pointing up. At t = 0.5 the ball reaches the top of its trajectory and starts going down. At t = 1 it passes the initial point and its velocity equals the initial velocity but points in the opposite direction.
There is one open question about the throwing force, though. What causes it to decrease? There are plenty examples of forces that decrease with distance. For example, the closer two magnets are, the stronger they attract or repel each other. Gravity and electric force between two charges also decrease with distance. But a ball that is thrown directly up initially moves away, but then moves closer and passes through the point where it started its movement. Our formula shows that throwing force decreases with time, not distance. Well, I heard that in Einstein's theory of relativity time and distance are parts of the same space–time continuum. But relativistic effects only becomes noticeable when the velocity is close to the speed of light...
To be honest, I have no idea what I am talking about.
reiterating the statement by johnsomeone ...
ignoring air resistance, the only force (therefore, the net force) acting on a projectile is the force of gravity, weight = mg , acting in the downward direction.
go to the link and read the second paragraph ...
What is a Projectile?
Ha-ha, got you. Yes, this was a joke, just like the question was a joke. And if the question was not a joke, then this was a reasonable exploration of the most likely incorrect answer. As far as I remember, my physics teacher had to explain that there is no "throwing force" and movement by inertia is accounted for by the second law of Newton. Also, this was an invitation to think about what would happen if the second derivative in the Newton's law is replaced by the first derivative. The result would be a world without inertia.