Understanding Mgh = 1/2mv^2: Conditions for Equal and Unequal Energy on Slopes

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The discussion clarifies the conditions under which gravitational potential energy (mgh) equals kinetic energy (1/2mv^2) when an object descends a slope. It emphasizes that energy conservation principles apply when no non-conservative forces, like friction, are present, allowing for the equality of potential and kinetic energy. If an object is released with initial velocity, the total initial energy includes both potential and kinetic energy, complicating the relationship. In the presence of friction, the decrease in potential energy results in a smaller increase in kinetic energy, as some energy is lost to the surroundings. Overall, energy conservation holds true when friction is absent, but the relationship changes with its presence.
ZGMF - X20A
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I just want to know in what conditions does mgh = 1/2mv^2 when a guy goes down a slope and in what conditions does mgh =/= 1/2mv^2. Thanks in advance.
 
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according to COE, (initial)PE + KE = PE + KE(final) so i think that its better to use this instead of your equation.For example,if the object were to be released with some initial velocity,the KE(initial) would not be 0 so total initial energy is mgh+1/2mv^2 instead of mgh.Btw,it is more precise to write change of PE = -change of KE as this shows that energy is conserved.
 
And energy is conserved as long as the are no "non-conservative" forces- i.e. as long as there is no friction.
 
O.o
So when there is no friction P.E = K.E?
 
Yes, decrease in PE= increase in KE and vice versa. If there's friction, decrease in PE= increase in KE + energy 'lost' to surroundings, so increase in KE<decrease in PE.
 
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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