What is the largest mass that can be placed on the spring

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SUMMARY

The largest mass that can be placed on a spring with a force constant of 1.50 N/m and a maximum extension of 10 cm is 0.153 kg. The calculation involves applying Hooke's Law (F = -kx) to determine the force exerted by the spring at maximum extension. The correct approach is to analyze the system in static equilibrium rather than using energy equations, as the force exerted by the spring is variable with extension. A free body diagram can clarify the relationship between force and mass in this scenario.

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  • Understanding of Hooke's Law (F = -kx)
  • Basic principles of static equilibrium
  • Ability to draw and interpret free body diagrams
  • Knowledge of gravitational force (E = mg)
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Homework Statement


the spring in a typical hooke's law apparatus has a force constant of 1.50 N/m and a maximum extension of 10cm. what is the largest mass that can be placed on the spring without damaging it?


Homework Equations


what i believe to be relevant:
F=-kx
W=F"d
E=mg(h)


The Attempt at a Solution


F=1.5(0.10)
F=0.15

W=F"d
W=0.15(0.10)
W=0.015J

0.015=m(9.81)(0.10)
0.015/0.981=m
m=0.0153kg <----------- answer

now i have no idea how this can be wrong.. i have gone over it many times now.
my textbook has it as 0.153kg.
am i right on this one?
 
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jaron said:

Homework Statement


the spring in a typical hooke's law apparatus has a force constant of 1.50 N/m and a maximum extension of 10cm. what is the largest mass that can be placed on the spring without damaging it?


Homework Equations


what i believe to be relevant:
F=-kx
W=F"d
E=mg(h)


The Attempt at a Solution


F=1.5(0.10)
F=0.15

W=F"d
W=0.15(0.10)
W=0.015J

0.015=m(9.81)(0.10)
0.015/0.981=m
m=0.0153kg <----------- answer

now i have no idea how this can be wrong.. i have gone over it many times now.
my textbook has it as 0.153kg.
am i right on this one?

I was able to solve this by using a force analysis. You don't need to even use energy.

Try finding what mass puts the system in static equilibrium at the spring's maximum extension and solve for mass.

There might be a way to solve it with the energy equations, but I don't really see a need to go that route. The only thing I can see is if you go that direction, the force exerted by the spring isn't constant, it's a function of x. But like I said, you don't even need to go that route.

Draw a free body diagram and the answer should be obvious.
 

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