Calculating Total Energy in a Pulley System: Kinetic vs. Potential Energy

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In a pulley system with a hanging mass and a cart on an air track, the total energy is conserved, comprising both kinetic and gravitational potential energy. The kinetic energy is calculated using Ke = 1/2(m1 + m2)V², while the potential energy of the hanging mass is given by mgh. It's essential to consider the potential energy of the cart as well, as both components contribute to the system's total energy. To demonstrate energy conservation, calculate the total energy at the initial state and after the mass drops, comparing the two results. This approach will clarify how energy transitions between potential and kinetic forms without loss.
RobtheSlob
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I am having a lot of trouble with a lab problem where a hanging mass is attached via a pulley to a cart, the friction is assumed negligible as the cart is on an air track. The cart is released and the hanging mass drops, pulling the cart along. I can't seem to find the right equations to include the cart into the system and compare the kinetic energy to the potential energy.
For the kinetic energy part I am using this equation:Ke= \frac{1}{2}(m_1 + m_2)V^2
 
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That's good. Keep in mind that the mass has gravitational potential energy equal to mgh. The total energy of the system should be conserved (assuming no friction) throughout its motion. What specifically are you trying to determine?
 
The object is to compare the two forms of energy and show how no energy is lost in the conversion from potential to kinetic. When I've worked the numbers I can't seem to figure out how to factor in the movement of the cart due to the potential energy of the hanging mass and if I don't it doesn't seem to come out right.
 
The cart also has potential energy (I forgot about that). You are trying to prove that total energy of the system (kinetic + gravitational potential) is constant. You can find the potential energy using the equation (I don't think you can do it experimentally), then use the formula you stated earlier for kinetic energy to find that (also making use of some data collected in the lab). Calculate total energy for the initial state of the system, and for after the mass is allow to drop, then compare.
 
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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