Conservation of energy of two blocks

AI Thread Summary
The discussion revolves around a physics problem involving two blocks connected to a spring, focusing on the conservation of energy. The participants calculated the combined kinetic energy of the blocks after one block falls a distance of 0.090 m, arriving at 2.7 J for part A and 1.8 J for part B. A key challenge identified was determining the maximum distance the hanging block falls before stopping, with the correct answer being 0.39 m. Participants sought clarification on how to connect gravitational potential energy with kinetic energy and spring potential energy in their calculations. The conversation emphasizes the importance of understanding energy conservation principles in solving the problem.
iamkristing
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[SOLVED] conservation of energy

Homework Statement


Two blocks, of masses M = 1.9 kg and 2M are connected to a spring of spring constant k = 220 N/m that has one end fixed, as shown in the figure below. The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed.

(a) What is the combined kinetic energy of the two blocks when the hanging block has fallen a distance of 0.090 m?
J

(b) What is the kinetic energy of the hanging block when it has fallen that 0.090 m?
J

(c) What maximum distance does the hanging block fall before momentarily stopping?
m
W0155-N.jpg



Homework Equations



k=(1/2)mv^2
u=mgh

ki + ui = kf + uf

The Attempt at a Solution



I solved A using u(grav) =kf +us and got 2.7 J
For B i got 1.8 J

C is where I have a problem. I know the kinetic energy must equal zero and somehow you find the height from u=mgh. I just can't seem to connect the two...
 
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iamkristing said:
C is where I have a problem. I know the kinetic energy must equal zero and somehow you find the height from u=mgh. I just can't seem to connect the two...
The change in energy must be zero. You start with gravitational PE, which gets converted to spring PE.
 


iamkristing said:


Homework Equations



k=(1/2)mv^2
u=mgh

ki + ui = kf + uf

The Attempt at a Solution



I solved A using u(grav) =kf +us and got 2.7 J
For B i got 1.8 J

C is where I have a problem. I know the kinetic energy must equal zero and somehow you find the height from u=mgh. I just can't seem to connect the two...


For part A, when you say you used Ugrav = final kinetic energy + Spring potential energy, could you please elaborate as to how that was accomplished? If final kinetic energy is 0.5*mass*velocity, how did you calculate velocity?

For part B, I run into the same roadblock, the veocity.

For C, I don't understand how the distance is 0. The answer is supposed to be 0.39m (I'm referring to the answsers in the back.

I apologise for asking so much, but I would really like to understand how this problem is solved. Thank you in advance.

EDIT: Just realized that the thread is about a year old! I apologise for that...
 
Last edited:


iceman2048 said:
For part A, when you say you used Ugrav = final kinetic energy + Spring potential energy, could you please elaborate as to how that was accomplished? If final kinetic energy is 0.5*mass*velocity, how did you calculate velocity?

For part B, I run into the same roadblock, the veocity.
For both A and B you are solving for the final kinetic energy.

For C, I don't understand how the distance is 0. The answer is supposed to be 0.39m (I'm referring to the answsers in the back.
The distance is not zero. (That earlier post of mine states that the change in energy is zero, not the distance. That's just a statement of energy conservation.)
EDIT: Just realized that the thread is about a year old! I apologise for that...
Yeah, it generally doesn't make sense to respond to an old post... but here we are!
 
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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