Conservartion of Energy - spring problem

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The discussion revolves around a physics problem involving a cookie sliding on a surface attached to a spring, with a given kinetic energy and friction acting on it. The initial kinetic energy of 19.5 J is converted into thermal energy due to friction and potential energy in the spring as it stretches. The correct calculation shows that the cookie will slide 0.30 m from the unstretched position before coming to rest, as the negative solution from the quadratic equation is not physically meaningful. The participants clarify the role of friction and the energy transformations involved in the system. The final conclusion emphasizes that only the positive displacement value is relevant for the problem.
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Homework Statement



A large cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 380 N/m; the other end of the spring is fixed in place. The cookie has a kinetic energy of 19.5 J as it passes through the position where the spring is unstretched. As the cookie slides, a frictional force of magnitude 10.0 N acts on it.

(a) How far will the cookie slide from the position where the spring is unstretched before coming momentarily to rest?


Homework Equations



Unsure


The Attempt at a Solution



All the spring potential energy is converted into kinetic energy when the spring reaches its equilibrium position.

Then, the spring undergoes kinetic friction, releasing thermal energy, which means all the energy at the equilibrium position will be converted into thermal energy.

Therefore:

20 = 10d
d = 2

Is this correct? I'm unsure, it seems like I'm missing something...
 
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Nynex said:
Is this correct? I'm unsure, it seems like I'm missing something...
No. You forgot about spring PE as the spring is stretched. Initially, at the unstretched position, there is only KE. That energy ends up as both thermal energy and spring PE.
 
Kinetic Energy = Thermal Energy + Potential Spring Energy
19.5 = 10d + 1/2kd2
0 = 180d2 + 10d -19.5
d = 0.30 m, -0.36 m

Is this correct? I think it makes sense...the spring isn't displaced as far to the right due to friction.
 
Nynex said:
Kinetic Energy = Thermal Energy + Potential Spring Energy
19.5 = 10d + 1/2kd2
0 = 180d2 + 10d -19.5
d = 0.30 m, -0.36 m

Is this correct? I think it makes sense...the spring isn't displaced as far to the right due to friction.
Yes, perfectly correct. Note: Only one of the solutions to that quadratic is physically meaningful for this problem. Which one?
 
It would be 0.30 m, it's initially compressed by 0.36 m to the left, the friction is what stops the spring from fully stretching out by 0.36 m to the right. Thanks for all the help!
 
Nynex said:
It would be 0.30 m,
Right.
it's initially compressed by 0.36 m to the left, the friction is what stops the spring from fully stretching out by 0.36 m to the right.
Not really. A negative value for d makes no sense, since that would mean that the friction actually increases the energy. So we just ignore that solution to the quadratic.
 
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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