Spring System with Masses: Force and Vertical Oscillations

  • Thread starter Thread starter Lewis
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
The discussion focuses on analyzing a spring-mass system where a box of mass M supports a block of mass m. When the system is displaced downward a distance d and released, the reaction force between the block and the box is determined as a function of time. The critical value of d is identified, which indicates when the block will just begin to leave the bottom of the box at the peak of its vertical oscillations. The analysis assumes no air resistance, simplifying the calculations. Understanding these dynamics is essential for solving the problem effectively.
Lewis
I'm having lots of trouble with this one, help would be much appreciated!

A spring of stiffness k supports a box of mass M, in which is placed a block of mass m. If the system is pulled downward a distance d from the equilibrium position and then released, find the force of reaction between the block and the bottom of the box as a function of time. For what value of d will the block just begin to leave the bottom of the box at the top of the verticle oscillations? Neglect any air resistance.
 
Physics news on Phys.org
Please read this thread

https://www.physicsforums.com/showthread.php?t=4825
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top