Spring potential energy comparison

AI Thread Summary
The discussion centers on comparing the potential energy of a spring at two different positions, A and B, after being stretched. The potential energy is calculated using the formula PE = 1/2 kx^2, where 'x' represents the distance the spring is stretched. It is noted that if the spring is stretched to double the distance at position B compared to position A, the potential energy at B will be four times greater due to the x^2 relationship in the equation. Participants are encouraged to clarify the problem statement and provide their reasoning for the solution. Understanding the implications of the spring's stretch distance is crucial for solving the problem accurately.
tourniquet63
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Homework Statement



A 0.10 m spring is stretched from equilibrium position to position A and then to position B as shown in the diagram below. Compared to the spring’s potential energy at A, what is its potential energy at B?

Homework Equations



PE = 1/2 kx^2

The Attempt at a Solution



Having trouble with this one. Is it twice as much or four times as much?
 

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Please retype the problem statement, the question is missing.
Please fill in section 2 also.
And state or show how you arrived at your answer.

Thanks, and welcome to these forums!
 
Ooops! thanks
 
In your relevant equation, what is the description of 'x'...what distance is it describing?
 
x is the distance the spring is compressed or stretched
 
tourniquet63 said:
x is the distance the spring is compressed or stretched
Ok so in the first part it is stretched a certain distance and in the second part it is stretched double that distance. Since the PE equation includes an x^2 term, if x doubles then the PE must be ___?___ times greater.
 
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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