Respuesta :
Answer:
Elastic potential energy, E = 3.26 J
Explanation:
It is given that,
Force constant of the spring, k = 5.2 N/m
Relaxed length of the spring, X = 2.45 m
When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m
To find,
The elastic potential energy stored in the spring.
Solution,
The extension in the length of the spring is given by :
[tex]x=x'-X[/tex]
[tex]x=3.57\ m-2.45\ m[/tex]
x = 1.12 m
The elastic potential energy of the spring is given by :
[tex]E=\dfrac{1}{2}kx^2[/tex]
[tex]E=\dfrac{1}{2}\times 5.2\times (1.12)^2[/tex]
E = 3.26 J
So, the elastic potential energy stored in the spring is 3.26 joules.
The Elastic Potential Energy stored in the spring mass system
= 3.26 Joule
The Elastic Potential Energy Stored (E) in the spring mass system is given by equation (1)
E = (1/2) [tex]\times K \times x^2[/tex]........(1)
Where K is the Force Constant = 5.2 N/m
and [tex]x[/tex] is the extension or compression of the spring.
Here as the spring length increases so [tex]x[/tex] = Final Length - Initial Length
= (3.57- 2.45) = 1.12 m
E = (1/2) [tex]\times[/tex]5.2 [tex]\times[/tex][tex]1.12^2[/tex] = 3.26144 [tex]\approx[/tex] 3.26 Joule
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