Springs / elastic potential energy / grav pot energy

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SUMMARY

The discussion focuses on calculating the spring constant, elastic potential energy, change in gravitational potential energy, and maximum height of an 8.00 kg box resting on a spring compressed by 7.0 cm. The spring constant is determined to be 11.2 N/cm using the formula F = kx, where F is the weight of the box. The elastic potential energy of the spring when compressed an additional 30.0 cm is calculated using the formula (1/2)kx², with total compression considered. The change in gravitational potential energy and maximum height are derived from the conservation of energy principles.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the formula for elastic potential energy: (1/2)kx²
  • Knowledge of gravitational potential energy calculations
  • Basic principles of energy conservation
NEXT STEPS
  • Learn how to derive the spring constant from force and displacement
  • Explore the relationship between elastic potential energy and gravitational potential energy
  • Study the concept of energy conservation in mechanical systems
  • Investigate real-world applications of springs in engineering and physics
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the principles of energy conservation and spring dynamics.

vtech
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You are given an 8.00 kg box resting on a spring. The spring is compressed 7.0 cm by the box.

(a) What is the spring constant? ____ N/cm

(b) The box is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
____ J

(c) What is the change in the gravitational potential energy of the box-Earth system when the box moves from the release point to its maximum height? ____ J

(d) What is that maximum height, measured from the release point?
_____ m



-------------------

I've got part A) which is 11.2 N/cm, I believe... but I'm having trouble getting past this...

help anyone?
 
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You should have an expression for the potential energy of a spring, something like [itex]\frac {k x^2} 2[/itex] you need to check that.

The change in gravitational potential energy is related to the change in height.


for d, consider that energy is conserved, at what point will the stored spring potiential energy be the same as the gravitational potential energy?
 
vtech said:
You are given an 8.00 kg box resting on a spring. The spring is compressed 7.0 cm by the box.

(a) What is the spring constant? ____ N/cm
You should know that "F= kx" where F is the force, x is the distance the spring is stretched or compressed and k is the spring constant: so k= F/x.
In this case, the force is the weight of the box. Since you were told that the answer must be in units of N/cm, you need to find the weight of the box in kg and then divide by 7 cm.

(b) The box is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
____ J
As Integral said, the potential energy of a spring compressed a distance x is (1/2)kx2. Note that the TOTAL compression here is 37 cm.

(c) What is the change in the gravitational potential energy of the box-Earth system when the box moves from the release point to its maximum height? ____ J
If the spring was compressed 30 cm from the point at which the box was at 'equilibrium' (where it was at rest originally), it should be easy to see that the boxes maximum height will be 30 cm above its equilibrium point. What is the change in gravitational potential energy of the box in moving up 60 cm?


(d) What is that maximum height, measured from the release point?
_____ m
Oops, I think I just handed you that one!
 

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