# Velocity->Force, concept problem

1. Jun 15, 2006

### seaglespn

Ok, so I have a problem in which an object is thrown tangentialy to a surface which makes 45o with the orizontal. So we know, that there is an MOMENT OF FORCE, an instant acceleration, which give the object the initial velocity : Vo.

The object is thrown in the upside direction(so is slowing down).

Now, we know the friction between the object and the surface (u=0.25).

The problem now is asking for the object's acceleration...

I am quite good at other kind of problems concerning Newton's Physics but this is blowing my mind :grumpy: , I don't know how to translate the initial velocity into a force, in order to do my basic rutine :).
Thx!

2. Jun 15, 2006

### arildno

Do NOT care about the acceleration period in which your object gained its initial velocity, due to the application of the force from your hand.
(You know way too little about that period).
Instead, focus on the period AFTER the force from your hand has stopped from being applied!

In particular:
What are now the forces acting upon the object?

3. Jun 15, 2006

### seaglespn

Well , there is an inertial force (gained from the initial velocity), a friction force and a tangential gravitational force.

Mathematicaly there must be Fi - mg*sin(alpha) - u*mg*cos(alpha) = mA
But I have problems trying to visualise how Fi can be calculated.
Must be something like m*Ao , but I have bigger problems if I think like that because mA MUST be negative in order to slow down the object, and if m*Ao>Ff+Gtg then the object (in my , obviously erronated universe :) ) dosn't slow down, but it gains speed :(.

4. Jun 15, 2006

### arildno

What the heck are you talking about??
Inertial force???? From initial velocity?
No such force exists.

5. Jun 15, 2006

### seaglespn

I must be wrong then , sorry .

I was thinking at that law of Newton which says :"objects in motion tend to remain in motion".

So what is it then?

6. Jun 15, 2006

### arildno

It doesn't say that:
Roughly it says that "an object NOT SUBJECTED TO EXTERNAL FORCES preserves its momentum". Inertial frames are those reference frames in which Newton's 1. law is valid.

So, your Fi doesn't exist; take care of the signs of your forces, the initial velocity won't appear in the expression for the initial acceleration.

7. Jun 15, 2006

### seaglespn

Ok, then can be one single acceleration counted?
Like : mA - Ff - Gtg = 0?

If this (^^^) is correct then A=(Ff + Gtg)/m =>
A= (u*mg*cos(alpha) + mg*sin(alpha))/m => A = u*g*cos(alpha) + g*sin(alpha) => A= g(u*cos(alpha) + sin(alpha)).

Is this( ^^^ ) correct ?

8. Jun 16, 2006

### arunbg

Yes, this is correct but only when the body is moving uphill ( both g and f act in the same direction ).