Question

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100kg of turkey. The slices of turkey are weighedon a plate of mass 0.400kg placed atop a vertical spring of negligible mass and force constant of 200N/m . The slices of turkey are dropped on the plate all at thesame time from a height of 0.250m . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You mayassume that the collision time is extremely small.

What is the amplitude A, of oscillations of the scale after the slices of turkey land on the plate?

Express your answer numerically in meters and take free-fall acceleration to be = 9.80m/s^2

I'm using the energy conservation and so far found v=2.2

Initial momentum=0.22 and final=.500kg

The final velocity is=0.44 and kinetic is=.048

I have taken K=.048=1/2kx^2-mgx

Anyways I'm stuck can anyone help?

Answer 1

kinetic energy=1/2kx^2-mgx
solve for x
.048=100x^2-9.8*.5*x
100x^2-4.9x-.048=0
x=0.057m=position of extremum=E


position of equillibrium
mg=kx
.5*9.8=200*x
x=.0245
A=.057-.0245=.0325m=32.5 cm