Simple harmonic motion frictionless block problem

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SUMMARY

The discussion focuses on a physics problem involving a block on a frictionless surface attached to a spring, oscillating with an angular frequency of 6.6 rad/s. To determine the minimum speed (v0) required for the block to knock over a bottle positioned 0.084 m away, participants emphasize the need to account for both kinetic energy and spring potential energy. The correct approach involves calculating the energy at maximum stretch when the block strikes the bottle and ensuring energy conservation between the launch and impact states. The final calculation for v0 is derived from the relationship between spring constant (k), mass (m), and displacement.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Knowledge of energy conservation in mechanical systems
  • Familiarity with spring constant (k) and mass (m) relationships
  • Ability to apply kinetic and potential energy equations
NEXT STEPS
  • Study the derivation of the spring constant (k) from angular frequency (ω) and mass (m)
  • Learn about energy conservation in oscillatory systems
  • Explore the equations of motion for simple harmonic oscillators
  • Investigate real-world applications of simple harmonic motion in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to energy conservation and simple harmonic motion.

triplel777
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Homework Statement


A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 6.6 rad/s. The drawing indicates the position of the block when the spring is unstrained. This position is labeled "x = 0 m." The drawing also shows a small bottle located 0.084 m to the right of this position. The block is pulled to the right, stretching the spring by 0.056 m, and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.



Homework Equations





The Attempt at a Solution



1/2mv_0^2=1/2kd^2
d=0.84-0.56=0.28
V_0=sqrt (k/m)*d
V_0= 6.6*0.28= 1.85

what am i doing wrong?
 
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1/2mv_0^2=1/2kd^2
d=0.84-0.56=0.28
Something wrong here. You need to say there is KE + spring energy initially and at least spring energy finally. Probably final KE as well since it has to hit fast enough to knock over a bottle. How fast is that?

Subtracting d values is not going to give you the same result as subtracting their squares.
 
You have to find the energy needed to stretch the spring so that the block reaches the bottle. This means the spring must stretch to .084 m.

After finding the k of the spring (what is the relationship between \omega and k and m?) write out the expression for the energy of the system (block+spring) at the point of maximum stretch when the bottle is struck. Write out the expression for energy of the system when the block is launched. How are the two expressions related? (hint:Assume that energy is conserved).

AM
 

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