Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
The spring constant of the spring will be "1000 N/m".
Given values are:
Force,
Length of spring,
As we know,
→ [tex]Applied \ Force = Spring \ constant\times Change \ in \ length[/tex]
or,
→ [tex]F = kx[/tex]
By substituting the values, we get
→ [tex]40=k\times (0.14-0.1)[/tex]
→ [tex]40= 0.04 \ k[/tex]
→ [tex]k = \frac{40}{0.04}[/tex]
[tex]= 1000 \ N/m[/tex] (Spring constant)
Thus the above answer is right.
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