Physics
WAEC 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
Correct Answer: Option A
Explanation
E = \(\frac{1}{2}Ke^2\)
= \(\frac{1}{2} \times 200(0.8)^2 = 64J\)
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