Work done by external force in a gravitational field.

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Homework Help Overview

The problem involves calculating the work done by an external agent to move a sphere of mass 212 g between two fixed spheres of masses 2.53 kg and 7.16 kg, which are separated by a distance of 1.56 m. The task requires understanding the gravitational interactions and the work-energy principle in a gravitational field.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for calculating the work done by the external force, questioning the signs in the equation based on the movement of the sphere relative to the two fixed masses.

Discussion Status

The discussion is ongoing, with participants exploring the reasoning behind the signs in the work equation. Some guidance has been offered regarding the interpretation of work done in relation to the gravitational forces involved, but no consensus has been reached.

Contextual Notes

Participants are grappling with the implications of the signs in the work equation and how they relate to the movement of the sphere between the two masses. There is an emphasis on understanding the physical meaning behind the calculations rather than just obtaining a numerical answer.

moatasim23
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Homework Statement



Spheres of masses m1 2.53kg and m27.16kg are fixed at a distance 1.56m apart centre to centre.A m3 212 g sphere is positioned 42cm from the 7.16 kg fromcentre of 7.16 kg sphere along the line of centres.How much work must be done by ext agent to move thesphere of 212 g sphere along the line of centres and place it 42 cm from the centre of sphere of mass 2.53 kg sphere?Assume the 212 g sphere to be moved bw the two spheres.

Homework Equations



Wext(External)=-Wg(gravitational)

The Attempt at a Solution


Wext=Gm3m2(1/.42-1/1.14)-Gm3m1(1/1.14-1/.42)
Which is not giving the desired result of 98.5pJ.
But if I take
Wext=Gm3m2(1/.42-1/1.14)+Gm3m1(1/1.14-1/.42) the answer gets right.But why should it be a positive sign?
 
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hi moatasim23! :smile:

(try using the X2 button just above the Reply box :wink:)
moatasim23 said:
Spheres of masses m1 2.53kg and m27.16kg are fixed at a distance 1.56m apart centre to centre.A m3 212 g sphere is positioned 42cm from the 7.16 kg fromcentre of 7.16 kg sphere along the line of centres.How much work must be done by ext agent to move thesphere of 212 g sphere along the line of centres and place it 42 cm from the centre of sphere of mass 2.53 kg sphere?Assume the 212 g sphere to be moved bw the two spheres.

so m3 starts 0.42 from one sphere and 1.14 from the other sphere, and finishes the other way round
… Wext=Gm3m2(1/.42-1/1.14)+Gm3m1(1/1.14-1/.42) the answer gets right.But why should it be a positive sign?

why not?

one is the work done moving it away from one sphere, the other is the work done moving it towards the other sphere, so the work done should have opposite signs, so your formula should have a + :confused:
 
tiny-tim said:
hi moatasim23! :smile:

(try using the X2 button just above the Reply box :wink:)


so m3 starts 0.42 from one sphere and 1.14 from the other sphere, and finishes the other way round


why not?

one is the work done moving it away from one sphere, the other is the work done moving it towards the other sphere, so the work done should have opposite signs, so your formula should have a + :confused:

That is what I am saying work done should have opposite signs.Taking both works to be positive does this means they hv opposite signs?How?
 
(just got up :zzz:)

not following you :confused:

your first bracket is negative, and your second bracket is positive, so they have opposite signs, as expected (and so you add them)
 

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