Block/Spring Problem (Kinetic Energy)

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SUMMARY

The discussion focuses on a physics problem involving a 13 kg block attached to a spring with a spring constant of 4.3 kN/m, which is stretched 13 cm and released. The kinetic energy calculations involve determining the energy at 4.8 cm from the release point, at the relaxed spring position, and the maximum kinetic energy achieved. The initial elastic potential energy is calculated as 36.34 J, and the conservation of energy principle is emphasized, factoring in a frictional force of 100 N that affects the block's motion.

PREREQUISITES
  • Understanding of elastic potential energy and its formula: ElasticPotentialEnergy = 1/2ky²
  • Knowledge of kinetic energy and its formula: K = 1/2mv²
  • Familiarity with Newton's second law: F = ma
  • Basic principles of conservation of energy in mechanical systems
NEXT STEPS
  • Calculate the kinetic energy of the block at 4.8 cm using energy conservation principles.
  • Determine the kinetic energy when the block passes through the relaxed spring position.
  • Analyze the effects of friction on the kinetic energy of the block.
  • Explore the relationship between spring constant and energy in oscillatory motion.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of spring-block systems and energy calculations.

lshine09
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Homework Statement



A 13 kg block on a horizontal surface is attached to a horizontal spring of spring constant k = 4.3 kN/m. The block is pulled to the right so that the spring is stretched 13 cm beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of 100 N.

(a) What is the kinetic energy of the block when it has moved 4.8 cm from its point of release?

(b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed?

(c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?f

Homework Equations



ElasticPotentialEnergy= 1/2ky^2
K=1/2mv^2
F=ma

The Attempt at a Solution



I honestly don't know where to begin. I used the numbers I was given to find the elastic potential energy at the beginning and came up with 36.34 J... and now I am not sure what to do.

(Us=1/2Ky^2
Us= (1/2)(4300N/m)(.13)^2
Us= (36.34J)
 
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Use the conservation of energy, remembering to account for the energy lost by friction.
 

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