# AS Physics Watch

Announcements

Page 1 of 1

Go to first unread

Skip to page:

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??

0

reply

Report

#2

(Original post by

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??

**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??

Contact force = weight

Larger weight = larger force needed

1

reply

(Original post by

Force needed = coefficient of friction x contact force

Contact force = weight

Larger weight = larger force needed

**Jackasnacks**)Force needed = coefficient of friction x contact force

Contact force = weight

Larger weight = larger force needed

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???

0

reply

Report

#4

**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??

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.

1

reply

(Original post by

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.

**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.

0

reply

Report

#6

(Original post by

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???

**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???

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

Posted from TSR Mobile

0

reply

Report

#7

(Original post by

Suppose the force is applied to the object by rolling an object down a hill?

**Jpw1097**)Suppose the force is applied to the object by rolling an object down a hill?

Posted from TSR Mobile

1

reply

(Original post by

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

Posted from TSR Mobile

**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

Posted from TSR Mobile

0

reply

Report

#9

(Original post by

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?

**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?

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.

Posted from TSR Mobile

0

reply

(Original post by

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.

Posted from TSR Mobile

**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.

Posted from TSR Mobile

So if you were to plot a graph of distance against mass, what would you expect?

0

reply

Report

#11

**Jpw1097**)

Suppose the force is applied to the object by rolling an object down a hill?

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:

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.

0

reply

**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

Posted from TSR Mobile

0

reply

Report

#13

(Original post by

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?

**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?

This going down a hill would be equal to the kinetic energy-the change in gpe.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top