Jpw1097
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Suppose you push an object with a given force and it moves a certain distance, why is that if you push a heavier object with the same force it will move less distance??
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Jackasnacks
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(Original post by Jpw1097)
Suppose you push an object with a given force and it moves a certain distance, why is that if you push a heavier object with the same force it will move less distance??
Force needed = coefficient of friction x contact force

Contact force = weight
Larger weight = larger force needed
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Jpw1097
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(Original post by Jackasnacks)
Force needed = coefficient of friction x contact force

Contact force = weight
Larger weight = larger force needed
So can the energy required to stop the object be seen as the amount of kinetic energy that must be dissipated??

If so does that mean that work done = braking force (frictional force) x braking distance
Therefore since the work done is the amount of energy needed to be dissipated:
KE(of object)=braking force x braking distance

Braking force = frictional force=μ x R

Since R=mg..does that mean that braking force =μ x mg

Therefore the amount of energy needed to be dissipated =μmg x distance

Therefore distance =work done/μmg???
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lerjj
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(Original post by Jpw1097)
Suppose you push an object with a given force and it moves a certain distance, why is that if you push a heavier object with the same force it will move less distance??
How long are you applying the force for? In a perfectly smooth frictionless world, the mass of the object only affects the acceleration (the more massive one accelerates less due to F=ma).

If we take into account friction, then a nice way to think about it is in terms of energy- if you give them both the same amount of kinetic energy to start with, then they will lose the energy through friction with the ground. The one with the highest friction will stop first and this will typically be the object with the greatest weight.
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Jpw1097
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(Original post by lerjj)
How long are you applying the force for? In a perfectly smooth frictionless world, the mass of the object only affects the acceleration (the more massive one accelerates less due to F=ma).

If we take into account friction, then a nice way to think about it is in terms of energy- if you give them both the same amount of kinetic energy to start with, then they will lose the energy through friction with the ground. The one with the highest friction will stop first and this will typically be the object with the greatest weight.
Suppose the force is applied to the object by rolling an object down a hill?
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Gizos
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(Original post by Jpw1097)
So can the energy required to stop the object be seen as the amount of kinetic energy that must be dissipated??

If so does that mean that work done = braking force (frictional force) x braking distance
Therefore since the work done is the amount of energy needed to be dissipated:
KE(of object)=braking force x braking distance

Braking force = frictional force=μ x R

Since R=mg..does that mean that braking force =μ x mg

Therefore the amount of energy needed to be dissipated =μmg x distance

Therefore distance =work done/μmg???
Yes. I think why you have said is right.

Basically
R=mg
F=μR
Therefore
F=μmg

And
Work done against friction=Fd
Work done against friction=μmgd

Rearranges to

D=work done/μmg


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Gizos
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(Original post by Jpw1097)
Suppose the force is applied to the object by rolling an object down a hill?
You have to then take into account the change in potential energy


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Jpw1097
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(Original post by Gizos)
Yes. I think why you have said is right.

Basically
R=mg
F=μR
Therefore
F=μmg

And
Work done against friction=Fd
Work done against friction=μmgd

Rearranges to

D=work done/μmg


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So providing the work which must be done is constant (which it is, since it is provided by rolling the same object down a hill at the same height/distance) then the distance is dependent on the weight of the object? Also, would the force applied be almost instantaneous?
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Gizos
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(Original post by Jpw1097)
So providing the work which must be done is constant (which it is, since it is provided by rolling the same object down a hill at the same height/distance) then the distance is dependent on the weight of the object? Also, would the force applied be almost instantaneous?
Yes but you'd have to remember that

R=mgcosx

With x the angle between the slope and horizontal.

Ok I think the force is applied till you overcome friction and start moving. (Ie you need a resultant force down the hill) but not entirely sure on this point.


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Jpw1097
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(Original post by Gizos)
Yes but you'd have to remember that

R=mgcosx

With x the angle between the slope and horizontal.

Ok I think the force is applied till you overcome friction and start moving. (Ie you need a resultant force down the hill) but not entirely sure on this point.


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Yes, I'm not interested in the actual weight component of the slope....I'm just assuming the kinetic energy converted to the object at the bottom of the slope is constant.

So if you were to plot a graph of distance against mass, what would you expect?
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lerjj
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(Original post by Jpw1097)
Suppose the force is applied to the object by rolling an object down a hill?
Then both keep on rolling until the hill runs out!

More qualitatively, you should probably do this using energy- the kinetic energy gained is equal to the change in GPE, but you've lost energy due to friction:

E_k=GPE-\mu N \times s

If you knew the slope of the hill and the co-efficient of friction you could calculate the resultant energy. The resultant velocity would be independent of mass, and since the heavier one would slow down faster at the bottom due to increased friction it would still travel the least distance.
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Jpw1097
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(Original post by Gizos)
Yes. I think why you have said is right.

Basically
R=mg
F=μR
Therefore
F=μmg

And
Work done against friction=Fd
Work done against friction=μmgd

Rearranges to

D=work done/μmg


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However, isn't the work done equal to the kinetic energy....therefore mv^2/2μmg = v^2/μg....so wouldn't the masses cancel out?
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Gizos
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(Original post by Jpw1097)
However, isn't the work done equal to the kinetic energy....therefore mv^2/2μmg = v^2/μg....so wouldn't the masses cancel out?
Sorry I shortened at the end. I worked out the work done due to friction.

This going down a hill would be equal to the kinetic energy-the change in gpe.
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Jpw1097
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That clears up a lot, thanks for the help
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