A spring with spring constants K when compressed by 1 cm, the potential energy stored is U.If it is further compressed by 3 cm, then its final potential energy is

16 U

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9 U

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8 U

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15 U

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Solution

The correct option is B $16 U$
Potential energy stored in a spring is given by:
$u = \frac{1}{2} k x^{2}$ [ $x :$ net compression]
when compressed by $x = 1$ cm, it stores energy $u = \frac{1}{2} k (1)^{2} = \frac{k}{2}$
when further compressed by $3$ cm, Net compression $= 3 + 1 = 4$ cm $= x^{'}$
Total PE stored
$u^{'} = \frac{1}{2} k x^{' 2} = \frac{1}{2} k (3 + 1)^{2} = 8 k$
Clearly, $\frac{u^{'}}{u} = 16$
$\Rightarrow u^{'} = 16 u$ .

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A spring with spring constants K when compressed by 1 cm, the potential energy stored is U.If it is further compressed by 3 cm, then its final potential energy is

The correct option is B 16UPotential energy stored in a spring is given by:u=12kx2 [x: net compression]when compressed by x=1cm, it stores energy u=12k(1)2=k2when further compressed by 3cm, Net compression =3+1=4cm=x′Total PE storedu′=12kx′2=12k(3+1)2=8kClearly, u′u=16⇒u′=16u.