Calculating Shock Absorber Energy Dissipation for Car Bounce

AI Thread Summary
To calculate the energy that shock absorbers must dissipate to dampen a car's bounce, one must consider the car's mass and initial velocity. The car, weighing 1240 kg and bouncing at 0.840 m/s, possesses kinetic energy that needs to be converted into thermal energy to stop its movement. While the spring equation Fs = -kx is relevant, shock absorbers function as damped springs, meaning they dissipate energy rather than just storing it. The key to solving the problem lies in calculating the initial kinetic energy of the car and understanding the role of the shock absorber in energy dissipation. Ultimately, the energy dissipated by the shock absorbers is equal to the initial kinetic energy of the car.
narutoish
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Homework Statement



How much energy must the shock absorbers of a 1240 kg car dissipate in order to damp a bounce that initially has a velocity of 0.840 m/s at the equilibrium position? Assume the car returns to its original vertical position.

Homework Equations



Fs = -kx

The spring equation is he only thing that comes even close to making sense.

The Attempt at a Solution



I honestly don't know how to even get started, do I use the spring equation or something else. We covered thermodynamics and waves recently so it has to be from there, I just can't find any equation that would work.

Thanks
 
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A shock absorber is more than just a spring. If it was just a perfect spring then it would not be very useful as you would just keep bouncing around. A shock absorber is basically a damped spring, it is a spring that is lossy in such a way that converts kinetic energy into thermal energy.

So in this case, we have a car that has started to bounce with a given mass and velocity. How much energy will you need to dissipate to stop it's vertical movement?
 
Kinetic energy, of course! Thank you.
 
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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