Physics
posted by Anonymous .
Need help with part b... I got a to be k = 94.985 N/m
I can't figure out how to go about doing part b..
A platform of mass 0.79 kg is supported on four springs (only two springs are shown in the picture, but there are four altogether). A chunk of modeling clay of mass 0.63 kg is held above the table and dropped so that it hits the table with a speed of v = 0.77 m/s.
The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6.5 cm below its original position.
a) What is the effective spring constant of all four springs taken together?
k = N/m *
94.985 OK
HELP: What is the net displacement Δx of the spring from its former equilibrium position? What force F was required to compress the springs? The effective spring constant is the ratio F/Δx.
b) With what amplitude does the platform oscillate immediately after the clay hits the platform?
A = cm
HELP: You might find it useful to find the speed of the table+clay system just after the collision, using momentum conservation.
HELP: Look at this from an energy point of view. Just after the collision, the total energy is partly potential (since the spring is now stretched from its new equilibrium position) and partly kinetic. When the spring is at its maximum displacement from equilibrium, all the energy is potential.
HELP: Remember that for a vertical spring, one can ignore gravity if one takes the equilibrium position as the origin. The equilibrium position is the place where the restoring force of the springs equals the weight of the clay.
HELP: Just after the collision, what is the potential energy stored in the spring? Remember that it is now displaced from its new equilbrium position. What is the kinetic energy? The sum of these is the total energy.
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