The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is
Question 1 Report
The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is
Answer Details
The energy stored in a spring can be calculated using the formula:
Energy (E) = 0.5 * k * x^2
where k is the stiffness constant of the spring and x is the amount it is extended or compressed from its rest position.
In this case, k = 2000 Nm^-1 and x = 0.04 m (4 cm converted to meters).
Substituting the values into the formula:
E = 0.5 * 2000 * 0.04^2
E = 0.5 * 2000 * 0.0016
E = 1.60 J
So, the energy stored in the spring when it is extended 4 cm is 1.60 J.
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