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For the system the Hamiltonian is : $$\notag H=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+\frac{K}{2}\left(x_2-x_1\right)^2 \ .$$ You write down the Hamilton equation of motion: \notag \begin{cases} \dot{x_1}=\frac{p_1}{m_1}\\ \dot{x_2}=\frac{p_2}{m_2}\\ \dot{p_1}=k(x_2-x_1)\\ \dot{p_2}=k(x_1-x_2) \end{cases} ...

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By using work energy theorem it can be solved. The velocity of the 2kg object till it reaches the 6kg object is given by $$\sqrt{2*g*5}=9.90m/s^2$$ apply the conservation of momentum for plastic impact. $$m_1u_1+m_2u_2=m_1m_2V$$ $$2*9.90+6*0=8*V$$ $$V=2.475m/s^2$$ work energy theorem $$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*x$$ $$x=1.801472656m$$ ...

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The total energy as a function of time is a constant and it is always equal to the potential energy in the point where $x=x_{max}$ and $v = 0$. The total energy is the work done to bring the free end of the spring from $0$ to $x_{max}$. So $E(t) = E = Work = kx_{max}^2/2 + ax_{max}^4/4$ You simply integrate $f(x)$ from $0$ to $x_{max}$

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Hint: Use $$m\ddot{x}=-kx-x^3 \\\ddot{x}=v\frac{dv}{dx} \\-\frac{kx^2}{2}-\frac{ax^4}{4}=\frac{m}{2}\left(\frac{dx}{dt}\right)^2$$ It will reduce to a form $$\frac{dx}{dt}=ix\sqrt{c^2+x^2}$$ This is a standard integral, and can be solved, then use $$U=-\int f(x) dx \\T=\frac{1}{2}m\dot{x}^2$$ Total energy $E=T+U\; .$

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The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both $m$ and the spring is characterized by constant $k$ the angular frequency of ...

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Yes it will. The spring behaves like friction here. When the first block is pushed, it transfers the energy to the spring which converts the kinetic energy to its potential energy. Once the second block overcomes its inertia, it will also start to move. Think of two blocks on a surface with friction without a spring in between. Pushing the first block ...

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where spring's energy goes It does not go. From the conservation of energy, $\frac12Kx_\text{max}^2=\frac12mv_\text{max}^2$, and $m = 0$ you get $v_\text{max}=\infty$. In conclusion, your spring will oscillate with $\text{Amplitude} = x_\text{max}\;,v_\text{max}=\infty \; \&\; \omega = \infty\; .$ For more information see: ...

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There are no ideal springs. Therefore the paradox with the infinite acceleration is not a physical one, but an artefact of the mathematical modelling. Conservation of energy does obviously not hold, when there are objects of mass zero in a system (because the kinetic energy will always be 0). So your setting simply does not fulfil the requirements for energy ...

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Normally it is said that the spring has no mass, but in problems where the spring is attached to bodies with larger mass compared to the spring mass. So you pull the string from the side where you have a body of mass $m$ (tha other side is attached to the wall). Then when you release this body with potential $\frac{1}{2} K x^2$ it will be converted in ...

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For part (1.) think about this: just as the spring comes to rest momentarily all of the energy in this system is stored in potential energy of the spring itself. First find the spring constant $k$ using Hooke's law. From there you can find the potential energy stored in the spring. You then know that just as the mass comes in contact with the spring the ...

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You can draw a schematic diagram to analyze it and suppose that elastic force is $F=-kx$ by Hooke's law and the gravity is $mg$.

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Thank you for your help. Here is a new model built by my friends. Even though this model is not perfect (don't consider the table and hand will also vibrate with phone), I think it is good enough for me to explain it. In this mode, phone equilibrium changes according to applied force. However, it is impossible to go below the table since the phone is ...

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The radius of gyration has a scalor value for each axis. For spring constant's dimensional analysis, it would have the dimensions corresponding to the unit's around it. So for example for hooks law: F=kX, where F is Force in Newtons and x is a distance in meters, then k, the spring constant, would have dimensions Newtons/meters. If you can be more specific ...

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If you use a constant force along the path, the spring will move past the position where $F=kx$, because it will reach that point at some speed. Thus it is incorrect to use the force method in the way you used it, because at maximal extension $v=0$ but $a\neq0$. The energy method as you used it will give the correct answer. If, instead, the force is used to ...

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The first method is giving the correct answer. In writing the work done by the force, you are assuming that the force $F$ itself is constant throughout the extension. However, this is not true. While extending the spring in a quasi-static way, the force $F$ must always match exactly the spring force at that time. This is needed so that at the end of the ...

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thanks for the reply. Here are two candidate answers I got from my friends and the other forum: Vibration amplitude decreases because the system damping factor is increased when the force of hand is applied Effective mass is increased when force is applied (because the table is vibrating also with the phone) Both of those factors are not considered on ...

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I think the reason is that you are just adding these two forces i.e. Force by hand and Force by the vibrating mechanism. But you have to understand that the hand comes in contact to the mobile surface only when the surface is going up(towards the hand) i.e. when the positive cycle of the vibration is taking place. When the vibration is moving on the other ...

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Note that the sign of the expression $$\frac12 \;k \; (x_f^2 - x_i^2)$$ depends on your choice of coordinate system. If you put $x_f=0$, then the result is negative; if you put $x_i=0$ then the result is positive. But the work done should be the same, regardless of the choice of coordinates. So what is happening? Simply, when we say $F = -kx$ we have ...

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This question arises because of a subtlety involved in the choice of variables and infinitesimals we use in the definite integration to find the work done by the force. If the block is at the coordinate $x$ and is moving towards the origin then the work done by the spring on it during the small interval of time in which the block travels a distance of $dl$ ...

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Notice, we know $$x_f<x_i$$ $$\iff x_f^2<x_i^2\ \ \ \ \ ( \forall \ \ \ x_i, \ x_f\ge0)$$ $$\iff x_f^2-x_i^2<0$$$$\iff k(x_f^2-x_i^2)<0 \ \ \ \ \ ( \forall \ \ \ k>0)$$ Now, the work done by the spring on the block is $$\int_{x_i}^{x_f}kx\ dx=\frac{1}{2}k(x_f^2-x_i^2)<0$$ Thus, the work done by the spring on the block is negative. It ...

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I'm not a physics guru (yet!), but here's my 50 cents: Since $\lambda$ represents complex vibration frequency, you're not interested in a negative real part of the root (as negative frequency isn't physical, or at least in the naive interpretation). So instead, you're taking the positive root, and then if $\lambda$ is a solution to the ODE, so is its ...

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If you draw force diagrams for each mass and then write Newton's 2nd law for each, you have two equations with three unkonwns: the elongation and the accelerations $a_1$ and $a_2$. To get one more key clue, we must note the following: When maximum elongation is reached it means that that spring doesn't stretch anymore. That means, it does not move relative ...

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The configuration of the system can be specified by two variables, the position of the end of the spring $z$, and the angle of the bar $\theta$. The kinetic energy can be written in terms of the time derivatives of these variables and the potential energy can be written in terms of these variables so you get your lagrangian. I think the take home message ...

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The formula for work done by a spring is: $$PE_{spring} = \frac{1}{2}k(\Delta x)^2$$ If I understand the question correctly, your $\Delta x$ is the 30 centimeters, or 0.3 meters. If you want to use the elastic potential energy approach, then you have to know the spring coefficient. The answers to your questions: 1) Yes, assuming the distance traveled ...

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I cannot seem to find the answer for my question. I know that the formula for work is $W=FD$. No, not really, not in most cases and not in this one. For a force vector $\vec{F}$ acting over an infinitesimal displacement vector $\vec{dx}$ the infinitesimal work done is $dW$: $$dW=\vec{F}\vec{dx}.$$ In your case both vectors are on the same line and ...

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In general, the force on the spring isn't going to be a helpful quantity. However, you can calculate how much energy is stored in the spring when it is compressed. From this energy, you should be able to calculate the energy of the block bullet system after impact, and then the velocity $v_2$. You should be able to do this using only conservation laws ...

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