Question about tension with ropes that have mass

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The discussion revolves around calculating the forces and tensions in a system of two blocks connected by ropes with mass, accelerated upward. The force required to accelerate the system is determined to be 32 N. The tension at the top of rope 1 is calculated as 19.2 N, while the tension at the bottom of rope 1 is 16 N. The tension at the top of rope 2 is found to be 3.2 N, with the assumption that there is no tension at the bottom of rope 2 due to it hanging freely. The calculations and assumptions presented are confirmed to be correct.
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Two 1.00 kg blocks are connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of .25 kilograms. The entire assembly is accelerated upward at 3.00 m/s^2 by force F.

a) what is F
b) what is the tension at the top end of rope 1?
c) what is the tension at bottom end of rope 1?
d) what is the tension at top end of rope 2?

My answers:

a) F=32 N

b) Tension at top of rope 1=19.2 N

c) Tension at bottom of rope 1= 16 N

d) Tension at top of rope 2= 3.2 N

I made the assumption that there was no tension at the bottom of rope 2 because there are no contact forces there, nothing is touching the bottom of the rope, it's just hanging. Everything depends on this assumption. Can you please tell me if my answers are correct? Thank you kindly for all your help.
 
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Looks good.
 
Thank you kindly for all your help!
 
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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