Friend pulls the spring without you looking

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The discussion revolves around a physics problem involving a mass attached to a spring, where the mass is pulled and released, reaching a speed of 3.375 m/s at the equilibrium position. The key focus is on applying the conservation of energy principle, equating elastic potential energy (Ee) and kinetic energy (Ek) to solve for the displacement (x) from the equilibrium position. The proposed equation is 1/2(k)(x)^2 = 1/2(m)(v)^2, which allows for the calculation of x. The participant expresses confusion about the energy states during the pulling process but confirms the approach to find the displacement is valid. This method effectively demonstrates the relationship between potential and kinetic energy in spring systems.
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Homework Statement


0.500kg mass resting on frictionless surface is attached to a horizontal spring with a spring constant of 45N/m. When you are not looking, your lab partner pulls the mass to one side and then releases it. When it passes the equlibrium position, its speed is 3.375m/s. How far from the equilibrium position did your lab partner pull the mass before releasing it?


Homework Equations


Elastic Potential Energy
Kinetic Energy


The Attempt at a Solution


What is confusing me is how I am going to apply the law of conservation of energy here. Could I say that whilst he is pulling the mass, only Ee is present and not Ek?

I was thinking I would do something like this to end up getting the "x" value needed:

1/2(k)(x)^2 = 1/2(m)(v)^2 <-- then solve for x, which is the amount of stretch or compression..
 
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aeromat said:
What is confusing me is how I am going to apply the law of conservation of energy here. Could I say that whilst he is pulling the mass, only Ee is present and not Ek?
Assume that when he releases the mass it starts from rest.

I was thinking I would do something like this to end up getting the "x" value needed:

1/2(k)(x)^2 = 1/2(m)(v)^2 <-- then solve for x, which is the amount of stretch or compression..
Sounds good to me.
 
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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