A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring
1 answer:
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
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Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
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And is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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Generally considering from initial position to a position of distance from the nucleus
Here represented the distance of the proton from the nucleus where the velocity is
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
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And is the change in electric potential energy from initial position to a position of distance
from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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Divide equation 2 by equation 1
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