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Byju's Answer

Standard XII

Physics

Work Done When Force Is Varying

Let a spring ...

Question

Let a spring with force constant k is compressed by $x_{1}$ . Stored potential energy of the spring $U_{1}$ = $\frac{1}{2}$ k $x^{2}$ . Let it be further compressed by $x_{2}$ . Let it be further compressed by $x_{2}$ . Change in potential energy of the spring

k ( - )

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k [ (() - )]

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Solution

The correct option is B
k [ (() - )]

$Δ$ U = $U_{f}$ - $U_{i}$

$U_{f} = \frac{1}{2} K (x_{1} + x_{2})^{2}$

$U_{i}$ = $\frac{1}{2}$ K ${x_{1}}^{2}$

$∴$ $Δ$ U= $\frac{1}{2}$ K $(x_{1} + x_{2})^{2}$ - $\frac{1}{2}$ k ${x_{1}}^{2}$

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Let a spring with force constant k is compressed by x1. Stored potential energy of the spring U1 = 12 k x2 . Let it be further compressed by x2. Let it be further compressed by x2 . Change in potential energy of the spring

The correct option is B k [ (() - )] ΔU = Uf - Ui Uf=12K(x1+x2)2 Ui = 12K x12 ∴ Δ U= 12 K (x1+x2)2 - 12 k x12

Let a spring with force constant k is compressed by $x_{1}$ . Stored potential energy of the spring $U_{1}$ = $\frac{1}{2}$ k $x^{2}$ . Let it be further compressed by $x_{2}$ . Let it be further compressed by $x_{2}$ . Change in potential energy of the spring

The correct option is B
k [ (() - )]

$Δ$ U = $U_{f}$ - $U_{i}$

$U_{f} = \frac{1}{2} K (x_{1} + x_{2})^{2}$

$U_{i}$ = $\frac{1}{2}$ K ${x_{1}}^{2}$

$∴$ $Δ$ U= $\frac{1}{2}$ K $(x_{1} + x_{2})^{2}$ - $\frac{1}{2}$ k ${x_{1}}^{2}$