Question

In an experiment, a student suspends a 700 g weight using a long spring mounted vertically to a lab stand, causing the spring to stretch as much as 18 cm after reaching equilibirum (at rest). She then pulls the weight at equilbrium position down for 8 cm and let go to set it into oscillation. Assuming this is an ideal system such that there is no lost (leaking) in energy such that the oscillate forever. What is the change in kinetic energy going from -A to A?

Answer 1

Given is:-

The weight of the particle is = 700gm or 0.7kg

the displacement in the spring to reach the equililbrium state is = 18cm or 0.18m

Thus at this point the force exerted by the spring is equal to the gravitational force, therefore

thus the value of spring constant is k=38.11 N/m

Now we know that according to the conservation of mechanical energy

where U is the potential energy and K is the kinetic energy of the spring system

here Kf-Ki is the difference between the final kinetic energy at -A and the initital kinetic energy at +A

we also know that the potential energy of spring system at distance x is given by

for this particular problem x is the distance from equilibrium position which = 0.08m = A

thus total potentiel energy is

by putting the values in above equation, we get

which is equal to the difference in the kinetic energies of the spring at -A and +A posititon

thus the difference between kinetic energies is 0.2439J