Understanding Work in Falling Bodies: Gravity, Energy, and Convention

[mpkannan]

When a body falls from a height (h) under gravity,the work involved is mgh. How do you describe this work? Is it 'work done by the force (gravity) on the body' or 'work done by the body'. The confusion is, if work is done on the body the body gains energy; but if work is done by the body the body loses energy. What is the standard convention?

 

[quantum123]

When a body falls from a height (h) under gravity,the work involved is mgh. How do you describe this work? Is it 'work done by the force (gravity) on the body' or 'work done by the body'. The confusion is, if work is done on the body the body gains energy; but if work is done by the body the body loses energy. What is the standard convention?

Not necessarily so. If the work is negative, the body loses energy.

 

[Doc Al]

When a body falls from a height (h) under gravity,the work involved is mgh. How do you describe this work? Is it 'work done by the force (gravity) on the body' or 'work done by the body'. The confusion is, if work is done on the body the body gains energy; but if work is done by the body the body loses energy. What is the standard convention?

Since gravity acts on the body, we're talking about work done on the body, not by the body. That work happens to be positive as the body falls, since the force and the displacement are in the same direction. (If the body were moving upward, the work done by gravity would be negative.)

 

[mpkannan]

Since gravity acts on the body, we're talking about work done on the body, not by the body. That work happens to be positive as the body falls, since the force and the displacement are in the same direction. (If the body were moving upward, the work done by gravity would be negative.)

Pl. see the following calcultions and advise me on the error involved:

(1) I consider a Cart coordinate system (drawn as per the right hand thumb rule) with +ve Z axis pointing upwards. Now, consider a body falling from z=h to z=0. The work involved is:

Integral (from h to 0) of [-mg dz] (-mg, because the force is acting downwards). The answer is mgh. Is this not the work done by the body, since it loses this much energy on fall? If so why it is not negative?

(2) If I raise the body from 0 to h, the work done on the the body is:

Integral (from 0 to h) of [mg dz] (here force applied is equal and opposite to gravity; so +mg). The asnwer is mgh. The body gains this much energy and stores it as its potential energy.

Why I am not getting opposite signs for the 2 types of work involved?

 

[quantum123]

(1) This is work done by the gravity on the body, not by the body. The body gains kinetic energy . Work done = gain in KE (this is work energy theorem) The integral is positive , if you have done your integration correctly.
(2) This is the wrong integral , as the gravity is still downward so it should be Integral (from 0 to h) -mg dz, and the integral this time is negative. So there is a loss of kinetic energy.

 

[Doc Al]

(1) I consider a Cart coordinate system (drawn as per the right hand thumb rule) with +ve Z axis pointing upwards. Now, consider a body falling from z=h to z=0. The work involved is:

Integral (from h to 0) of [-mg dz] (-mg, because the force is acting downwards). The answer is mgh. Is this not the work done by the body, since it loses this much energy on fall? If so why it is not negative?

No, it's not the work done by the body. Gravity (-mg) acts on the body.

(2) If I raise the body from 0 to h, the work done on the the body is:

Integral (from 0 to h) of [mg dz] (here force applied is equal and opposite to gravity; so +mg). The asnwer is mgh. The body gains this much energy and stores it as its potential energy.

Good.

Why I am not getting opposite signs for the 2 types of work involved?

Positive work was done in each case. In case 1, gravity did the work; in case 2, you did the work.

In case 2 the work done by gravity (not you) is negative, but you didn't calculate that.

 

[quantum123]

My apologies for the (2) as I have not noticed that you are calculating the work done by hand and not work done by gravity.
So yes, the work is integral from 0 to z (mg dz).
In both cases, the work is positive because they share a common thing:
The force is in the same direction as the displacement.