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Question about Work in physics

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DrDolittle

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Work

Consider a briefcase lying at rest on the ground. When i lift the briefcase to a table at height "h" i do positive work "mgh" and the gravity does negative work eventuating in zero work. This holds with the formula for work,that is, the difference in kinetic energy between 2 states is zero. When i bring the breifcase off the table i do negative work. What direction the gravity should act and why ?
As the net work done is agin zero in the second case the gravity should upwards. How ?

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drdolittle :)
 

jayc

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Re: Work

Work is defined as \[W = F\times d\]. When you are lifting up the brieffcase, assuming you do it at a constant velocity, \[F_{applied} = mg\] and the distance d = h. This value for work is positive because the force is applied in the same direction as the displacement, so the vectors are parallel. Gravity also applies a force of magnitude \[F_{g} = mg\] over distance h. However, the direction of the force applied and the displacement are opposite for gravity, so the vectors are antiparallel, giving the work done by gravity a negative value.

When you are bringing the briefcase back to the table, the opposite is true. The force you apply (again assuming constant velocity) is still \[F_{applied} = mg\] and the distance is still h, but now, you are applying an upward force and the displacement is downward, making the vectors antiparallel. This give you negative work done by you. Similarly, the work done by gravity is now positive because the force it applies is downward, and the displacement is also downward.
 

DrDolittle

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Re: Work

You haven't dealt with both the cases.
Case 1 : ground to table -- upward movement -- work done(positive(mgh)) -- gravity(negative(-mgh)) -- net work done -- 0

Case 2 -- table to ground -- downward mobvement -- work done(negative)--gravity(positive,that is upwards) -- net work done -- 0

My question is only with the second case. When the briefcase is brough down how could the gravity act upwards??

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drdolittle
 

jayc

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Re: Work

jayc said:
Similarly, the work done by gravity is now positive because the force it applies is downward, and the displacement is also downward.

I did deal with both cases. When bringing the briefcase downward, gravity is acting in the same direction as the displacement, while you are acting in the opposite direction.

Gravity always acts downward. You are applying an upward force to oppose the gravitational force in order to maintain constant velocity over the descent of the briefcase. In review, please note these points:

1) The briefcase has a displacement in the "down" direction.
2) Gravity always applys a force in the "down" direction.
3) You apply a force in the "up" direction to make net force = 0, maintaining contant velocity.
4) The force you apply is opposite in the opposite direction of displacement, making work done by you negative.
5) The gravitational force is applied in the same direction of displacement, making work done by gravity positive.

Perhaps I can make it even more clear by introducing a simple notation. Let's call the up direction positive (+) and down direction negative (-).

We agree that that \[W = F\times d\]. Note the following for the case of moving the briefcase down:

1) \[d = -h\]
2) \[F_{g} = -mg\]
3) \[F_{applied} = mg\]

Therefore,

4) \[W_{g} = (F_{g}) (d) = (-mg)(-h) = mgh\]
5) \[W_{you} = (F_{applied})(d) = (mg)(-h) = -mgh\]

Hope this clears things up.
 

DrDolittle

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Re: Work

I got mixed up with the direction of displacement. Thats very clear.

Regards
drdolittle :)

Added after 20 minutes:

Lets hash out some insights and have some healthy discussions.

When the briefcase is moved up and down along a fixed direction we can say it is moving with constant velocity. When it takes a haphazard path,still the speed remaining constant, the velocity is not constant......agreed?
 

jayc

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Re: Work

Actually, the definition of work we have is a simplified version for a special case where the force applied is constant over the displacement. The complete definition is actually a line integral:

\[W = \int\limits_a^b\overline{F} \cdot \overline{ds}\]

where a and b are the initial and final positions, respectively, of the mass acted on.

As you can see, if the displacement does not follow a straight path, we must calculate the integral based on the force (as a function of displacement) and the path taken.

You can also refer to this link:
Work as an integral
 

DrDolittle

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Re: Work

The incremental element ds will be pointing in different directions at different instances which is accounted for velocity. the velocity is still not considered constant, ain't it ?

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drdolittle :)
 

jayc

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Re: Work

Of course, if the direction of incremental displacement changes, the velocity cannot be constant. However, we only stated the constant velocity condition in your original question to simplify the problem (by concluding that force must be constant). In general, the velocity of the object being acted on is irrelavant. All you need is the path it travels and the force applied to it at each point.
 

DrDolittle

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Re: Work

what can you say about perceived gravity, that is, why do we perceive it that way :) ?

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drdolittle
 

sidra

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Re: Work

actully when the briefcase is comming towards the ground , u are not doing work on it...the ground is doing work on it....infact u r stopping the case to fall free on the ground and still u are doing work on the case in the up direction...so in both cases u are doing work away from the ground....may be u will understand from this.....sidra_maheen84@yahoo.com
 

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