Mastering Physics: 'weighing lunch': spring oscillation

In summary, a group of friends stop at a deli for lunch and order a sandwich with 0.100 kg of turkey. The turkey slices are weighed on a plate with a mass of 0.400 kg placed on a vertical spring with a force constant of 200 N/m. The slices are dropped from a height of 0.250 m, causing the plate to undergo vertical simple harmonic motion. The amplitude of oscillations A of the scale after the slices land on the plate is calculated using the equations F=-kx, deltaW=0.5kx^2, deltaW=0.5mv^2, and mv+mv=(m+m)vf (inelastic). The resulting answer is 0
  • #1
louza8
58
0

Homework Statement



For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 {\rm kg}of turkey. The slices of turkey are weighed on a plate of mass 0.400 {\rm kg} placed atop a vertical spring of negligible mass and force constant of 200 {\rm N/m}. The slices of turkey are dropped on the plate all at the same time from a height of 0.250 {\rm 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.

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{\rm m/s}^2.

Homework Equations



F=-kx
deltaW=0.5kx^2
deltaW=0.5mv^2
mv+mv=(m+m)vf (inelastic)

The Attempt at a Solution



m=0.1kg
m2=0.4k
k=200N/m
h=0.25

Initial equilibrium with 0.4kg plate:

ma=kx
(0.4*9.81)/200 = x
x=0.0196m

New equilibrium position with turkey

Need velocity of turkey+plate

v^2=u^2 + 2ash

=sqrt(2*9.81*0.25)
v=2.21472m/s

momentum after inelastic collision

mv=(m+m)v_f

(0.1*2.215)/0.5=0.442m/s

equate energies

0.5(m+m)v^2 + 0.5kx^2 = 0.5kA^2

sqrt([0.5(0.5)0.442^2 + 0.5(200)(0.0196^2)]/100)

A=0.029462m

so this says kinetic energy of turkey+plate + energy in initial displacement = final energy

answer is 0.027 I'm not sure where I stuffed up, i think in the initial displacement maybe?

no turkies were harmed in writing this question
 
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  • #2
With out going through it step by step, it looks right and your answer is close enough to reason that, try carrying the values to an extra digit to remove possible rounding errors.
 

Related to Mastering Physics: 'weighing lunch': spring oscillation

1. How does a spring oscillation help in weighing objects?

A spring oscillation is a type of simple harmonic motion where an object attached to a spring moves back and forth in a predictable pattern. By measuring the frequency and amplitude of this oscillation, we can calculate the force exerted by the object on the spring, which is directly related to its weight.

2. What is the equation used for calculating weight using spring oscillation?

The equation used for calculating weight using spring oscillation is F = -kx, where F is the force exerted by the object on the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position. By rearranging this equation, we can solve for the weight of the object, W = -kx.

3. How does the spring constant affect the accuracy of the weight measurement?

The spring constant determines the stiffness of the spring and how much it will stretch or compress when a force is applied to it. A higher spring constant means that the spring will be less affected by the weight of the object, resulting in a more accurate weight measurement. However, the spring constant must be known or calibrated beforehand in order for the measurement to be accurate.

4. Can spring oscillation be used to weigh any type of object?

Spring oscillation can be used to weigh any type of object, as long as it can be attached to the spring and its weight does not exceed the maximum weight that the spring can support without breaking or permanently deforming. However, the accuracy of the measurement may vary depending on the shape and size of the object.

5. What are some limitations of using spring oscillation for weight measurement?

One limitation of using spring oscillation for weight measurement is that it requires knowledge of the spring constant, which may vary depending on the quality and condition of the spring. Another limitation is that the object's weight must be within a certain range to avoid damaging the spring. Additionally, external factors such as air resistance and friction can affect the accuracy of the measurement.

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