Double-Checking Elastic Energy and Gravitational Potential | Dec. 14 Deadline

AI Thread Summary
The discussion focuses on verifying calculations for elastic energy and gravitational potential energy before a December 14 deadline. Participants inquire about the equations used, specifically the spring potential energy formula Us = 0.5kx^2 and gravitational potential energy formula Ug = mgh. Concerns are raised about the starting point measurements, particularly why the spring's unstretched value is lower than the initial position. It is noted that gravitational potential should decrease as elastic energy increases, indicating a potential error in the calculations. The urgency of the deadline emphasizes the need for accurate verification of these energy concepts.
Raza
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I want to make sure if I have done the Eleastic Energy part and also the Gravitational Potential correctly. Please respond fast. handing it on Dec.14
ElasticEnergy.jpg


Thank YOU :smile:
 
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hmm...which equations did u use? for the spring potential did u use:
Us=.5kx^2

did u use Ug=mgh grav. potential?

this may be just b/c i don't understand ur table...y is the spring unstretched value lower than ur starting point?
 
Last edited:
I used Ug=mgh to find the gravititional potential.

This is a picture to give you an idea of how it is.
ElasticEnergy2.jpg


The starting point is 7cm plus the extra 2cm. We had to hold it there. After that, I let of the weight and when it bouncing, i had to quickly measure the lowest point. Now, I know I did something wrong. because gravititional potential should be decreasing as the elastic energy is increasing.
 
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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