How Much Kinetic Energy Does a Ball Gain When Air Resistance Is Considered?

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When a ball drops and loses 30 J of gravitational potential energy while considering air resistance, it gains less than 30 J of kinetic energy. This is due to air resistance converting some of the potential energy into other forms, such as thermal energy, rather than fully converting it to kinetic energy. The consensus among participants is that the correct answer is less than 30 J, as air resistance acts as a non-conservative force. If no air resistance were present, the ball would gain exactly 30 J of kinetic energy. Therefore, the presence of air resistance significantly impacts the energy conversion process.
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Homework Statement


A ball drops some distance and loses 30 J of gravitational potential energy. Do not ignore air resistance. How much kinetic energy did the ball gain?

A. More than 30J
B. Exactly 30J
C. Less than 30J
D. Cannot be determined from the information given.


Homework Equations



GPE = MGH
KE = .5mv^2


The Attempt at a Solution



I think the answer should be less than 30J, because those are conservative forces, but friction due to air resistance takes some potential energy away. My lab partner says it should be more than 30J, for the same reason. Help?
 
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i agree that it should be C because the air resistance would restrict it from gaining the full kinetic energy from it's potential energy.
 
C. Less than 30.

If all the potential energy was transferred to kinetic it would be exactly 30 J,, but since there is air resistance, energy is changed into other forms other than kinetic, leaving less than 30.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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