Given some body with a constant force F acting on it upwards

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When a constant upward force F acts on a body subject to gravitational force, the net force must account for gravity to apply the Work-Kinetic Energy theorem correctly. The change in kinetic energy is determined by the work done by the force F minus the work done against gravity. If F is slightly larger than the gravitational force mg, the body accelerates upward, while if F is significantly larger, gravity's effects can be neglected. The discussion emphasizes the importance of considering both forces to accurately assess energy changes. Ultimately, the correct statement aligns with the application of the Work-Kinetic Energy theorem, confirming the relationship between work, kinetic energy, and gravitational potential energy.
xmjolx
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Given some body with a constant force F acting on it upwards
(The body is subject to the gravitational force due to the earth)
What's the correct statement?

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Or:

gif.gif


K - Kinetic Energy
 
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Think about how the potential energy changes with \Delta{x}. The change in total energy, which includes both kinetic and potential energy, will equal the total amount of work done on the body.

You might also find it helpful to consider the cases: F is only very slightly larger than mg; and F is so large relative to mg that we can ignore the effects of gravity.
 
what if F is small?
how come gravity has no affect on the change in kinetic energy?
 
xmjolx said:
Given some body with a constant force F acting on it upwards
(The body is subject to the gravitational force due to the earth)
What's the correct statement?

gif.gif


Or:

gif.gif


K - Kinetic Energy
If you want to apply the "Work - KE" theorem, you must use the net force on the body. That net force must include gravity.
 
Doc Al said:
If you want to apply the "Work - KE" theorem, you must use the net force on the body. That net force must include gravity.

So
gif.gif
is the right statement?
 
xmjolx said:
So
gif.gif
is the right statement?
Yes.
 
Thank you very much! :)
i knew it was the right answer but i had an argument with my high school teacher(she claims to have Ph.D in physics) about that.
 
Multiply out the rhs, then
\Delta K = F \Delta x - mg \Delta x.
Identifying the terms on the right as work or energy:
\Delta K = work\;done\;by\;F - \Delta (grav\;potential\;energy).
 
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