Finding the distance an object will travel

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SUMMARY

This discussion focuses on calculating the distance an object will travel when struck by a force on a horizontal surface. Key factors include the mass of the object, the frictional force, and the initial velocity, which is necessary for determining the object's acceleration and distance traveled. The conversation emphasizes the importance of understanding impulse and momentum, as well as the relationship between force, mass, and acceleration using Newton's second law (F=ma). The final distance can be derived using the work-energy principle, specifically the equation E = F*s, where E is the kinetic energy and s is the distance.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of impulse and momentum concepts
  • Familiarity with the work-energy principle
  • Basic understanding of friction and its effects on motion
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  • Research the impulse-momentum theorem and its applications
  • Learn how to calculate frictional force and its impact on motion
  • Study the work-energy principle in detail
  • Explore kinematic equations for constant acceleration scenarios
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Physics students, engineers, and anyone involved in mechanics or experimental physics who seeks to understand the dynamics of motion and the effects of forces on objects.

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Hi. I have a question about finding the distance an object will travel along a horizontal surface. If I have a force striking an object directly in the horizontal direction, how would I go about finding how far it will travel? I already have found the frictional force acting on the object being struck, I have the mass of the object being struck, and the force striking the object. Just for clarification, this object will be sliding across a surface, not rolling. Also, there will not be, (or shouldn't be, hopefully) any movement in the vertical direction. If anyone can help me with this, it would be greatly appreciated.
 
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A force striking an object doesn’t mean anything. A force can act on an object for a certain period of time, due to which the momentum or velocity of the body is changed.

If the frictional force is constant, then force/mass will give the deceleration. Then just apply the formula for const accn, if you know the initial speed. The final speed is zero.
 
It's simple if consider the friction coeficient constant. The initial velocity is known (you've got to calculate it somehow), you can calculate the dynamic energy of the object: E =0.5mv^2. Then the distance is derived from the work done by friction force:
E = A = F*s.
 
Shooting star, if I use the formula for constant acceleration, then I will end up with this:
(V^2-V_0^2)/(2*a) = \Delta x
If I use this equation with no acceleration, then I won't be able to get a distance because you can't divide by zero. Or, is the acceleration not zero?
 
Last edited:
That's right. If the body comes to rest, then V=0, so you can find delta_x if you know the initial velo. (Remember, 'a' is -ve.)
 
what is "e" in the equation a = -ve?

Oh, also, I only know the force striking the object, I don't know the initial velocity.
 
Last edited:
"-ve" is short for negative.

As I have said earlier, "a force striking an object" is meaningless. You must know the duration for which it acted on the body, or some other information.
 
Ok, the force acted on the body for an instant; it struck the object, and the object started moving. It was similar to a hockey stick hitting a puck.
 
Then you have to give some info about what the impact did to the body -- chnage in momentum, energy, whatever...
 
  • #10
Are you sure this isn't an impulse momentum problem, from there you can find the constant acceleration.
 
  • #11
It's not really a problem, I'm doing an experiment where I hit an object with a hammer to see how far it travels. I have to determine whether the mass of the object affects how far it travels. Clearly it does, but I need some kind of relationship to prove it.
 
  • #12
F=ma-u, (F+u)/m equals net acelleration=a. now what is needed is the time the object travels when coming to a complete stop.
 

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