Increasing GPE & KE: Can Force x Distance Be Avoided?

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SUMMARY

To increase gravitational potential energy (GPE) or kinetic energy (KE) of an object, work must be done, which involves applying a force over a distance. When a ball is kicked upwards, a significant force is applied over a brief distance, resulting in an increase in momentum. This interaction confirms that even when a force is applied for a short duration, it still contributes to the work done, as defined by the equation W=Fd. Therefore, while the force and distance may vary, work is always a necessary component in energy transfer.

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  • Understanding of gravitational potential energy (GPE)
  • Knowledge of kinetic energy (KE)
  • Familiarity with the work-energy principle
  • Basic physics concepts of force and distance
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  • Explore the relationship between force, distance, and energy transfer
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If i wanted to increase the gravitational potential energy of a ball do I always have to apply a force times a distance? In this case the force would be the weight of the ball and the distance would be the height.
What if I kicked the ball upwards? I wouldn't be applying a force times a distance , but I would be giving it momentum. So my question is, I don't always have to apply a force times a distance to increase the GPE or KE of an object do I?
 
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You always have to do work, which means applying a force over a distance. When you kick the ball, you're applying a large force over a small distance, but that distance is not quite zero. Your toe is in contact with the ball and applying some force to it for maybe one-tenth of a second, your toe is moving during that one-tenth second so covers some distance, that distance is the ##d## in ##W=Fd##.
 
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Nugatory said:
You always have to do work, which means applying a force over a distance. When you kick the ball, you're applying a large force over a small distance, but that distance is not quite zero. Your toe is in contact with the ball and applying some force to it for maybe one-tenth of a second, your toe is moving during that one-tenth second so covers some distance, that distance is the ##d## in ##W=Fd##.
Excellent! Thanks a lot!
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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