To confirm my answer wheter is right or wrong

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The discussion focuses on deriving the expression for the maximum kinetic energy of a mass attached to a vertical spring undergoing simple harmonic motion. The correct formula for maximum kinetic energy is confirmed as Kmax = π^2 * m * a^2 * k. There is a clarification regarding the use of angular frequency in the kinetic energy formula. The initial suggestion of using 1/2*m*v^2 was corrected to reflect the proper relationship involving spring constant and amplitude. The solution is validated, emphasizing the importance of accurate formula application in physics.
Sanosuke Sagara
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A mass m at the end of a vertical spring constant k vibrates vertically with simple harmonic motion of amplitude a.Find the expression for the maximum kinetic energy of the mass in terms of m,k and a.


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i think your maximum kinetic energy should be 1/2*m*v^2 = 1/2*m*w^4*x^2 in stead of 1/2*m*w^2*x^2
 
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Your solution is correct. The maximum kinetic energy of the mass can be expressed as:

Kmax = 1/2 * m * (2π * a * √(k/m))^2

or in simplified form as:

Kmax = π^2 * m * a^2 * k

Great job on solving this problem!
 
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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