Question on: WAEC Physics - 2004
An elastic string of force constant 200N m-1 is stretched through 0.8m within its elastic limit. Calculate the energy stored in the string
A
64.0J
B
80.0J
C
128.0J
D
160.0J
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Correct Option: A
E = \(\frac{1}{2}Ke^2\)
= \(\frac{1}{2} \times 200(0.8)^2 = 64J\)
= \(\frac{1}{2} \times 200(0.8)^2 = 64J\)
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