Velocity->Force, concept problem

  • Thread starter Thread starter seaglespn
  • Start date Start date
  • Tags Tags
    Concept
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving an object thrown tangentially to a surface inclined at 45 degrees to the horizontal. The original poster attempts to determine the object's acceleration, given its initial velocity and the friction coefficient between the object and the surface.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the forces acting on the object after the initial force has ceased, questioning the role of initial velocity and how it translates into force. There is discussion about the correct interpretation of Newton's laws and the concept of inertial forces.

Discussion Status

Some participants provide guidance on focusing on the forces acting on the object after the initial force is removed. There is a progression towards a mathematical expression for acceleration, with some participants confirming the correctness of the derived equation under specific conditions.

Contextual Notes

Participants are navigating misunderstandings regarding inertial forces and the application of Newton's laws, indicating a need for clarification on these concepts as they relate to the problem at hand.

seaglespn
Messages
26
Reaction score
0
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 , I don't know how to translate the initial velocity into a force, in order to do my basic rutine :).
Thx!
 
Physics news on Phys.org
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?
 
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 :(.
 
What the heck are you talking about??
Inertial force? From initial velocity?
No such force exists.
 
arildno said:
What the heck are you talking about??
Inertial force? From initial velocity?
No such force exists.

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

Similar threads

Replies
5
Views
1K
  • · Replies 47 ·
2
Replies
47
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 56 ·
2
Replies
56
Views
5K
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K