Work Done by Gravity: Formula & Considerations

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SUMMARY

The work done by gravity on an object falling from a height can be calculated using the formula W = F * s, where F is the gravitational force and s is the distance fallen. In this case, for a ball of mass 3 kg falling from a height of 20 m with g = 9.8 m/s², the work done by gravity is 588 J. However, when considering air resistance, the net work done is reduced, as air friction performs negative work against the gravitational force. Thus, while gravity does 588 J of work, the total work done is less due to the opposing force of air resistance.

PREREQUISITES
  • Understanding of Newton's Laws of Gravitation
  • Basic knowledge of work and energy concepts in physics
  • Familiarity with the formula W = F * s
  • Awareness of the effects of air resistance on falling objects
NEXT STEPS
  • Study the impact of air resistance on falling objects in physics
  • Learn about the principles of work and energy in classical mechanics
  • Explore advanced applications of Newton's Laws in real-world scenarios
  • Investigate the concept of negative work and its implications in physics
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of falling objects and the effects of air resistance on work done by gravity.

alingy1
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Let's say a ball falls from a height of 20m from the ground. g=9,8m/s^2. m=3kg.
Consider resistance of air in a qualitative way.

What's the formula for the work done by gravity?

So, I just thought about this question. I wonder if W=F*s should be applied. Does gravity still do 588J of work? Or should we take into account that F is smaller because of air friction? All this concept of work is very new to me and I googled for similar questions but couldn't find any web page alluding to my problem.
 
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If considering air resistance, then the net work is not the same as the work done by gravity, which is the same regardless of air resistance.

Your notion of W=F*S works if you treat gravity as an approximately constant force, but that approximation becomes poor as S gets larger. This all comes from the fact that the gravitational force on Earth or any other planet is actually given by Newton's Laws of Gravitation, which creates a more difficult work calculation.

You should also note that work is not simply F*S.

See: https://en.wikipedia.org/wiki/Work_(physics )
 
Last edited by a moderator:
alingy1 said:
Let's say a ball falls from a height of 20m from the ground. g=9,8m/s^2. m=3kg.
Consider resistance of air in a qualitative way.

What's the formula for the work done by gravity?

So, I just thought about this question. I wonder if W=F*s should be applied. Does gravity still do 588J of work? Or should we take into account that F is smaller because of air friction? All this concept of work is very new to me and I googled for similar questions but couldn't find any web page alluding to my problem.

The force due to gravity is the same, the distance fallen is the same, so the Work done by gravity should still be the same.
HOWEVER: the resistance of air will be doing some work too.
 
Okay, so, the total work done (not only by g) will be smaller than the work that should have been done without air. The air does negative work and g, the same work.
 
alingy1 said:
Okay, so, the total work done (not only by g) will be smaller than the work that should have been done without air. The air does negative work and g, the same work.

Correct. It is similar to pushing a box with friction opposing the push. You do work with your own force F_p as you move the block, while the ground also does work in the OTHER direction (different signs) with the frictional force F_f doing work as the block moves.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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