elastic energy help?
2017-01-17 7:31 pm
The elastic potential energy stored in a wire is 1 J for an extension of 1 mm. Find the extra energy stored in the wire if the length is further increased by 1 mm.
回答 (1)
2017-01-17 7:45 pm
Elastic potential energy, P.E. = (1/2) k x²
where k is the elastic constant, and x is the displacement (length of extension).
When x = 1 mm :
(1 J) = (1/2) k (1 mm)²
k = 2 J/mm²
Hence, P.E. = (1/2) * (2 J/mm²) * x²
When x = (1 + 1) mm = 2 mm :
P.E. = (1/2) * (2 J/mm²) * (2 mm)² = 4 J
The extra energy stored in the wire = (4 - 1) J = 3 J
where k is the elastic constant, and x is the displacement (length of extension).
When x = 1 mm :
(1 J) = (1/2) k (1 mm)²
k = 2 J/mm²
Hence, P.E. = (1/2) * (2 J/mm²) * x²
When x = (1 + 1) mm = 2 mm :
P.E. = (1/2) * (2 J/mm²) * (2 mm)² = 4 J
The extra energy stored in the wire = (4 - 1) J = 3 J
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