NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

Two springs A and B have force constants \[{{k}_{A}}\] and \[{{k}_{B}}\] such that \[{{k}_{B}}=2\,{{k}_{A}}\]. The four ends of the springs are stretched by the same force, If energy stored in spring A is E, then energy stored in spring B is: [AIPMT 2001]

A) E/2

B) 2 E

C) E

D) 4 E

Correct Answer: A

Solution :

Key Idea: When a string is stretched, then work done in stretching it through a distance x is the potential energy stored in it.

Potential energy stored = Work done is stretching

or \[U=\frac{1}{2}k\,{{x}^{2}}\]

also \[F=k\,x\]

or \[x=\frac{F}{x}\]

So, \[U=\frac{1}{2}k\,{{\left( \frac{F}{k} \right)}^{2}}\]

i.e., \[U\propto \,\,\frac{1}{k}\] (for cons \[\tan \,t\] force)

\[\therefore \] \[\frac{{{U}_{B}}}{{{U}_{A}}}=\frac{{{k}_{A}}}{{{k}_{B}}}\]

but \[{{k}_{B}}=2{{k}_{A}}\]

\[\therefore \] \[{{U}_{B}}={{U}_{A}}\times \frac{{{k}_{A}}}{2\,{{k}_{A}}}=\frac{{{U}_{A}}}{2}=\frac{E}{2}\]