Understanding the Relationship between Kinetic and Potential Energy in Freefall

AI Thread Summary
When a ball is dropped from a height, its potential energy (PE) decreases as it falls, while its kinetic energy (KE) increases. The increase in KE occurs in equal amounts over equal distances rather than equal time intervals, as the ball accelerates due to gravity. The conservation of mechanical energy principle states that the total mechanical energy (Emech) remains constant, meaning the loss in PE equals the gain in KE. To fully understand this relationship, it's essential to apply the equations for KE and PE. Properly using these equations will clarify the connection between the changes in energy types during freefall.
bohobelle
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Homework Statement


You drop a ball from a high balcony and it falls freely. Does the ball's kinetic energy increase by equal amounts in equal time intervals, or by equal amounts in equal distances? Explain.

Homework Equations


KE = 1/2mv^2
PE = mgy
Conservation of energy: Emech = KE + PE; Esys = Emech + Eth

The Attempt at a Solution


I think that it increases by equal amounts in equal distances, seeing as potential energy has to do with height. I'm just not sure how to explain it?
Thanks :)
 
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bohobelle said:

Homework Statement


You drop a ball from a high balcony and it falls freely. Does the ball's kinetic energy increase by equal amounts in equal time intervals, or by equal amounts in equal distances? Explain.

Homework Equations


KE = 1/2mv^2
PE = mgy
Conservation of energy: Emech = KE + PE; Esys = Emech + Eth

The Attempt at a Solution


I think that it increases by equal amounts in equal distances, seeing as potential energy has to do with height. I'm just not sure how to explain it?
Thanks :)
You are correct. You should write out the conservation of mechanical energy equation which you haven't done yet, that relates the change in KE with the change in PE.
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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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