The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm...
PHYSICS
WAEC 2017
The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is
- A. 0.16J
- B. 1.60J
- C. 16.00J
- D. 160.00J
Correct Answer: B. 1.60J
Explanation
The spring constant = the measure of stiffness = 2000\(Nm^{-1}\)
Energy stored in a spring = Work done to stretch it through a distance = change in potential energy
= \(\frac{kx^{2}}{2}\) = \(\frac{2000 \times (0.04)^{2}}{2}\) (converting 4cm to metres)
= \(\frac{3.2}{2} = 1.60J\)
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