Potential energy and kinetic energy confusion

AI Thread Summary
The discussion revolves around the confusion between potential energy (P.E.) and kinetic energy (K.E.) in a rollercoaster scenario. The total energy of the system is given as 4000 joules, with the mass of the trolley at 20 kg and gravity at 10 m/s². The key point is that as potential energy decreases, kinetic energy increases, maintaining the conservation of total energy. To find the kinetic energy, one must subtract the potential energy from the total energy, which clarifies why the teacher used this method. Understanding this relationship is crucial for solving physics problems effectively.
roxas
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okay this question is confusing like everything i currently know about p.e and k.e.
okay so the maximum energy of a rollercoaster is 4000j. the mass of the trolly is 20kg, and the acceleration due to gravity is 10m/s^2(dunnoe where's the squared button on the keyboard) okay so i am supposed to calculate the velocity at this point. my teacher gave the working for this answer and he minus the potential energy from the total energy to find the k.e, and this is where i got confused. isn't the potential energy lossed =the gain of K.E? why did he minus the P.E from the T.E? help! my physics practical is a day away and i can't afford to be confuzzled.:cry: :cry:

by the way i didnt clarify this wif my teacher as i was daydreaming in class and just copied down the answer. thx:biggrin:
 
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The total energy consists of kinetic and potential energy, as you know already. Since total energy is conserved, maximizing kinetic energy must minimize potential energy, and vice versa. So, an object has its maximal kinetic energy when its potential energy is minimal. The same holds for maximal potential energy.
 
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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