Respuesta :
Answer:
The elastic potential energy stored in the compressed piece is 5.47 J
Explanation:
Given;
elastic potential energy stored, E = 14 J
initial compression, x = 0.240m
The elastic potential energy stored in a spring is given as;
E = ¹/₂Kx²
where;
E is elastic potential energy
K is the spring constant
x is the compression
K = 2E / x²
K = (2*14) / (0.24²)
K = 486.11 N/m
When the spring is cut into three equal pieces, and one of pieces is then compressed by 0.150m, the energy stored in this piece is calculated as;
E = ¹/₂Kx²
E = ¹/₂(486.11)(0.15²)
E = 5.47 J
Therefore, the energy stored in one of the cut pieces compressed by 0.150m is 5.47 J
Answer:
5.47 J
Explanation:
From Hook's law,
Energy stored in a elastic material = 1/2ke²
E = 1/2ke²................. Equation 1
Where k = spring constant of the material, e = extension.
make k the subject of the equation
k = 2E/e²................... Equation 2
Given: E = 14 J, e = 0.24 m
Substitute into equation 2
k = 2(14)/0.24²
k = 28/0.0576
k = 486.11 N/m.
The elastic potential energy stored in the compressed piece is
Given: e = 0.15 m
E = 1/2(466.11)(0.15²)
E = 5.47 J