How Fast Will the Block Move After Being Released from the Spring?

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SUMMARY

The discussion centers on calculating the velocity of a 1.54-kg block after being released from a spring with a force constant of 1.47E+4 N/m, compressed by 0.100 m. The correct approach involves using the potential energy stored in the spring, given by the formula \(\frac{1}{2}kx^2\), and equating it to the kinetic energy of the block, \(\frac{1}{2}mv^2\). This method effectively accounts for the energy transfer and allows for the accurate calculation of the block's velocity after release.

PREREQUISITES
  • Understanding of Hooke's Law (F = kx)
  • Knowledge of kinetic energy formula (\(KE = \frac{1}{2}mv^2\))
  • Familiarity with potential energy concepts in springs
  • Basic principles of energy conservation
NEXT STEPS
  • Calculate potential energy in springs using \(\frac{1}{2}kx^2\)
  • Explore energy conservation principles in mechanical systems
  • Learn about dynamics of mass-spring systems
  • Investigate the effects of varying spring constants on block velocity
USEFUL FOR

Students in physics, engineers working with mechanical systems, and anyone interested in understanding energy transfer in spring-block dynamics.

volraithe
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A 1.54-kg block is held against a spring of force constant 1.47E+4 N/m, compressing it a distance of 0.100 m. How fast is the block moving after it is released and the spring pushes it away?

i thought u just use the formulas that F=kx and then plug that into F=ma, but i am not getting the right answer... anyone have any other suggestions as to what path I should take?
thank you.
 
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volraithe said:
A 1.54-kg block is held against a spring of force constant 1.47E+4 N/m, compressing it a distance of 0.100 m. How fast is the block moving after it is released and the spring pushes it away?

i thought u just use the formulas that F=kx and then plug that into F=ma, but i am not getting the right answer... anyone have any other suggestions as to what path I should take?
thank you.

Doing that will only give you the instantaneous acceleration. The problem with this is that the force will change continuously as the mass is pushed with the spring; this results in a constantly changing acceleration and we can't use our kinematics very easily. We want to use something else.

A very simple method for this problem would be to use the fact that the potential energy of the spring is \frac{{kx^2 }}{2}. We know that upon release, the spring is going to transfer all of it's energy into the mass. We of course know the energy of a mass as <br /> \frac{{mv^2 }}{2}. Equating both sides allows you to solve for the velocity of the mass.
 

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