WAEC - Physics (2004 - No. 10)
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
64.0J
80.0J
128.0J
160.0J
Объяснение
E = \(\frac{1}{2}Ke^2\)
= \(\frac{1}{2} \times 200(0.8)^2 = 64J\)
= \(\frac{1}{2} \times 200(0.8)^2 = 64J\)