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__Question__- A stretched elastic band of mass 0.67g is released so that its initial velocuty is horizontal, and its initial position is 9.6 com above the floor. What was the elastic potential energy stored in the stretched band if, it has a horizontal displacement of 3.6 m from its initial position?

__Solution__

Find v

At finish (h = 0)

Et = Ek + Eg

Et = 0.000335v^2

At Start (h = 0.096m)

Et = Ek + Eg

Et = 0.000335v^2 + 0.0006309792

by Law of Conservation of Energy (LCE)

Et = 0.000335v^2

0.000335v^2 = 0.000335v^2 + 0.0006309792

so when I got there, I was stuck and thought I went down the wrong track

But my method was going to be to find the velocity so I could find Ek which equals Ee because:

W = Ek

W = 1/2 k x^2

Ee = 1/2 x^2

so

Ee = Ek