Weighing Lunch For lunch you and your friends decide to stop at the nearest deli

Question

Weighing Lunch

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

Part A

What is the amplitude of oscillations A 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 g = 9.80 m/s2 .

A = ?? m

Part B

What is the period of oscillation T of the scale?

Express your answer numerically in seconds.

T = ?? s

Explanation / Answer

a)

here

The scale spring had already been compressed before the meat was dropped on it. Its initial compression was such that F = mg = kx, or
0.4kg * 9.8m/s² = 200N/m * x
x = 0.0196 m
U = ½kx² = ½ * 200N/m * (0.0196m)² = 0.04 J
The new equilibrium point is lower after the meat is added:
0.1kg * 9.8m/s² = 200N/m * x
x = 0.0049 m
Let's pick the point of maximum compression of the scale as our reference point.
initial E = PE + U = mg(0.25m + 0.0049m + A) + 0.04J
PE = 0.1kg * 9.8m/s² * (0.2549m + A) + 0.04J = 0.2898 + 0.98A
where A is the amplitude of motion.

At maximum compression,
E = U = ½k(0.0196m + 0.0049m + A)² = ½ * 200N/m * (0.0245 + A)²
E = U = 100 * (0.0245² + 0.049A + A²) = 0.06 + 4.9A + 100A²

These two energy levels are equal:
0.2898 + 0.98A = 0.06 + 4.9A + 100A²
100A² + 3.92A - 0.2298 = 0
A = -0.07 m, 0.032 m take positive root as amplitude

b)

T = 2/ = 2(m/k) = 2(0.5kg / 200kg/s²) = 0.31 s

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