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Standard XII

Physics

Work Energy Theorem Application

A spring obey...

Question

A spring obeying the linear law $F = - K x$ is first compressed by $10 c m$ and the work done is $W_{1}$ . Next it is compressed by another $10 c m$ , the work done now is $W_{2}$ , then $W_{1} : W_{2}$ is:

1 : 3

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3 : 1

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1 : 6

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6 : 1

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Solution

The correct option is A $1 : 3$
$W_{1} = Δ (P . E)$
$= \frac{1}{2} K (x {_{2}}^{2} - x {_{1}}^{2})$
$= \frac{1}{2} K (10^{2} - 0^{2})$
Similarly, $W_{2} = \frac{1}{2} K (20^{2} - 10^{2})$
So, $W_{1} : W_{2} = 100 : 300$ $= 1 : 3$

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A spring obeying the linear law F=−Kx is first compressed by 10cm and the work done is W1. Next it is compressed by another 10cm, the work done now is W2, then W1:W2 is:

The correct option is A 1:3W1=Δ(P.E)=12K(x22−x12)=12K(102−02)Similarly, W2=12K(202−102)So, W1:W2=100:300 =1:3

A spring obeying the linear law $F = - K x$ is first compressed by $10 c m$ and the work done is $W_{1}$ . Next it is compressed by another $10 c m$ , the work done now is $W_{2}$ , then $W_{1} : W_{2}$ is:

The correct option is A $1 : 3$
$W_{1} = Δ (P . E)$
$= \frac{1}{2} K (x {_{2}}^{2} - x {_{1}}^{2})$
$= \frac{1}{2} K (10^{2} - 0^{2})$
Similarly, $W_{2} = \frac{1}{2} K (20^{2} - 10^{2})$
So, $W_{1} : W_{2} = 100 : 300$ $= 1 : 3$