A weight of mass 2.00 kg hangs on a lineair spring with spring constant k= 3.25N/m^2. The weight is held at height y=0 so that the spring is not extended and exerts no force on the weight. When the weight falls, there is a damping force is proportional to the velocity. The initial velocity of the weight is zero. At time t=0, the weight is released. It falls and executes damped oscillations. Eventually, it comes to rest. For this situation, Newtons second law can be written as : m*(d^2*y/d*t^2)=-(dy/dt)-(k*y)-m*g(adsbygoogle = window.adsbygoogle || []).push({});

Here -(dy/dt) is the damping force, -ky is the force exerted by the spring and -mg is the force of gravity

Question: let y_final be the height of the weight after it stops moving..what is y_final?

I dont know how to solve this....pls who can give me hints?

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# Homework Help: Need help with this

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