Maximum speed of an object after a spring

AI Thread Summary
The discussion focuses on calculating the maximum speed and acceleration of a mass connected to a spring on a frictionless track. The mass is 534 g, the spring constant is 6 N/m, and the initial displacement is 2 cm. Participants discuss the relevant formulas, including potential energy in the spring, PE=(1/2)(k)(x²), and the need for proper calculations to determine maximum speed and acceleration. One user initially calculated a maximum speed of 0.037 m/s, which was questioned as being too low, prompting requests for detailed calculations. Accurate application of the formulas is essential for determining the correct values.
liamtcarroll
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A 534 g mass is connected to a light spring of
force constant 6 N/m that is free to oscillate
on a horizontal, frictionless track. The mass
is displaced 2 cm from the equilibrium point
and released from rest.

What is the maximum speed of the mass?

What is the maximum acceleration of the mass?

I found the period to be 1.87

Now I don't know what formula to use for the maximum speed, or to find the maximum acceleration.
 
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Do you know the formula for the energy stored in a spring? What will the stored energy be when the mass is at max speed?
 
is it Hooke's law?
 
wait is it PE=(1/2)(k)(x²)?
 
i got .037 m/s for my speed, does that sound right?
 
liamtcarroll said:
i got .037 m/s for my speed, does that sound right?
Sounds too low. Pls post your working.
 
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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