Work done by external force in a gravitational field.

[moatasim23]

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?

 

[tiny-tim]

hi moatasim23!

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

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 +

 

[moatasim23]

hi moatasim23!

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


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 +

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?

 

[tiny-tim]

(just got up :zzz:)

not following you

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

 

[Nickie]

I would like to clarify that the work done by an external force in a gravitational field is the negative of the change in gravitational potential energy of the system. In this case, we are considering the work done by an external agent to move the 212 g sphere from its initial position to its final position between the two fixed spheres.

The gravitational potential energy of a system of two masses separated by a distance r is given by the equation U = -Gm1m2/r, where G is the universal gravitational constant. Therefore, the work done by an external force to move the 212 g sphere from its initial position to its final position can be calculated as follows:

Wext = -ΔU = -[(-Gm3m2/r2) + (-Gm3m1/r1)]

Where r2 is the distance between the 212 g sphere and the 7.16 kg sphere, and r1 is the distance between the 212 g sphere and the 2.53 kg sphere.

Substituting the values given in the problem, we get:

Wext = -[(-6.67e-11*0.212*7.16)/(0.42) + (-6.67e-11*0.212*2.53)/(1.14)]

Wext = -(-9.96e-11 + -3.38e-11) = 1.33e-10 J = 133 pJ

Therefore, the work done by the external agent to move the 212 g sphere is 133 pJ, which is the desired result.

In conclusion, the negative sign in the equation for work done by an external force in a gravitational field represents the fact that the external force is doing work against the gravitational force of the system. This is why it is important to use the negative sign in the calculation to get the correct result.