What Is the Speed of a Mass Displaced from Equilibrium on a Spring?

AI Thread Summary
To find the speed of a mass displaced from equilibrium on a spring, the discussion highlights using energy conservation principles. The mass of 0.340 kg is attached to a spring with a force constant of 13.0 N/m and is initially displaced 0.250 m. The correct approach involves calculating potential energy at the maximum displacement and equating it to kinetic energy at the desired position. The user struggles with the calculations, providing various incorrect speed answers like 0.87 m/s and 1.54 m/s. The suggestion to utilize energy conservation is emphasized as a more effective method for solving the problem.
nlkush
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Homework Statement



A 0.340kg- mass is attached to a spring with a force constant of 13.0 N/m.
If the mass is displaced 0.250m from equilibrium and released, what is its speed when it is 0.140m from equilibrium?
in m/s
for some reason cannot get the correct answer.
 
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Show what you did.
 
omega=square roote of k\m
T=1.47
k=13.0
not sure of amplitude
eqn v=-Aomega*sin(omegat)
 
answers were like 0.87 m/s 0.5981m/s
1.54m\s
i cannot seem to get it correct
 
Why don't you use energy conservation.
 
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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