The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
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Answer:
19.68 × 10⁻³ m
Explanation:
Given;
Original Length, L₁ = 41.0 m
Temperature Change, ΔT = 40.0°C
Thermal Linear expansion of steel is given to be, ∝ = 12 × 10⁻⁶ /°C
Generally, Linear expansivity is expressed as;
∝ = ΔL / L₁ΔT
Where
∝ is the Linear expansivity
ΔL is the change in length, L₂ - L₁
L₂ is the final length
L₁ is the original length
ΔT is the change in temperature θ₂ - θ₁ (Final Temperature - Initial Temperature)
From equation of linear expansivity
ΔL = ∝L₁ΔT
ΔL = 12 × 10⁻⁶ /°C × 41.0 m × 40.0 °C
ΔL = 19.68 × 10⁻³ m
ΔL = 19.68 mm