Is a 15.0-g Ball from a Spring Gun Safe Without Goggles for Children?

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SUMMARY

The discussion centers on calculating the maximum speed of a 15.0-g ball shot from a spring gun with a spring constant of 600 N/m, compressed by 5.00 cm. The user attempts to apply the conservation of energy principle, equating the work done by external forces to the changes in kinetic and potential energies. The user’s calculations differ from the book's answer, prompting a request for clarification on the gravitational potential energy change. The recommendation on the necessity of goggles for children is also sought based on the calculated speed.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with spring mechanics and Hooke's Law
  • Knowledge of kinetic and potential energy equations
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the maximum speed of a projectile using the formula for kinetic energy
  • Explore the implications of spring constant variations on projectile speed
  • Investigate the effects of gravitational potential energy on vertical motion
  • Review safety guidelines for children’s toys involving projectile motion
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Physics students, educators designing educational games, and safety regulators assessing children's toys for potential hazards.

knowNothing23
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Homework Statement


Conservation of energy
You are designing a game for small children and want to see if the
ball’s maximum speed is sufficient to require the use of goggles. In your game, a
15.0-g ball is to be shot from a spring gun whose spring has a force constant of
600 N/m. The spring will be compressed 5.00 cm when in use. How fast will the
ball be moving as it leaves the gun and how high will the ball go if the gun is
aimed vertically upward? What would be your recommendation on the use of
goggles?

I have set the gravitational potential energy zero at the mouth of the gun; therefore, the initial gravitational potential energy is negative.

Could anyone help?
Thank you.


Homework Equations





The Attempt at a Solution


My velocity's answer is different than the one from the book.

Wext = ΔK + ΔU g + ΔU s = 0
or, because Ki = Us,f = Ug,f = 0, **"f" as in final**

(1/2)m(vf)^(2) - mgx − (1/2)k (x)^(2) = 0
 
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knowNothing23 said:
I have set the gravitational potential energy zero at the mouth of the gun; therefore, the initial gravitational potential energy is negative.
OK, but what's the change in gravitational PE?
 
-+-= +.

Thank you.
 

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