Two springs of force constant $$2000$$ $$N/m$$ and $$4000$$ $$N/m$$ are stretched by same force. The ratio of their respective potential energies is

Question

Two springs of force constant $$2000$$ $$N/m$$ and $$4000$$ $$N/m$$ are stretched by same force. The ratio of their respective potential energies  is

Infinity Learn · Accepted Answer

$$PE$$ of mass attached to a spring is $$PE$$$$=12kx2$$, where $$k=$$ force constant, $$x=$$ displacement restoring force of a spring is $$F=-$$$$kx$$,                                                ⇒ $$x=-Fk$$,substitute in $$PE$$ equation                                                   $$PE$$$$=12kFk2$$                                               ⇒$$PE$$$$=12kF2k2$$                                               ⇒$$PE$$ $$=12F2k$$                                               ⇒$$PE$$α$$1k$$ since stretched by same force                                               ⇒$$PE1PE2=k2k1=40002000=21$$  obtained after substituting the given values of force constant $$k1$$,$$k2$$

Two springs of force constant 2000 N/m and 4000 N/m are stretched by same force. The ratio of their respective potential energies is

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