Solved: Physics 20-1 Energy - Mass on a Spring

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The discussion revolves around a physics problem involving a mass on a spring with a spring constant of 700 N/m, compressed by 16 cm, and a maximum speed of 12 m/s. The user applies the conservation of energy principle, equating potential energy (Ep) and kinetic energy (Ek), to solve for mass (m), arriving at a value of 0.12 kg. Doubts arise regarding the correctness of the solution, especially since classmates obtained different results. The user also questions whether the orientation of the spring (vertical or horizontal) affects the answer. Clarification on the spring's orientation could impact the final result.
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Homework Statement


This is a question from a test i had today..i want to know if I did it right or not.

A mass on a spring with spring constant 700 N/m is compressed 16 cm. If the maximum value of the speed is 12 m/s, what is the mass of the mass in kg?


Homework Equations



Ep = 1/2kx^2, Ek = 1/2mv^2


The Attempt at a Solution



so conservation of energy plays a role in this...

Vmax = 12 m/s
k = 700 N/m
x = 0.16 m
m = ?

Ep = Ek
1/2kx^2 = 1/2mv^2
m = (1/2kx^2) / (1/2v^2)
m = 0.12 kg.

im really doubting myself right now.
 
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Looks ok to me, but then again it is 1:30 A.M and I am borderline asleep...
 
lol. well thanks i guess. but a whole bunch of other people in my class got something else. it's wierd.
 
Did the problem statement tell that whether the spring was vertical or horizonal?
if it's vertical ,the answer will be different.
 
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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