Question

1) A spring of constant K = 100N/m is extended from an initially released position to an extension of 10cm .What is the
(1) Ws
(2) ∆Us


2) Refer figure..​

Answer 1

Explanation:

$\huge\bf{ \mid{ \overline{ \underline{{Question:-} \mid}}}}$

1) A spring of constant K = 100N/m is extended from an initially released position to an extension of 10cm .What is the

(1) Work done by spring

(2) ∆U of spring

$\huge\bf{ \mid{ \overline{ \underline{{Answer:-} \mid}}}}$

$\large \red{Given \: that} \\ \\ K = 100N {m}^{ - 1} \\ \\ x1 = 0m \\ \\ x2 = 10cm = 0.1m$

$\\ \huge \purple{Formula} \\ \\ \Delta U = - W_{s} \\ \\ W_{s} = \frac{1}{2} \times K \times ( {x_{1}}^{2} - {x_{2}}^{2} )$

$\implies W_{s} = \frac{1}{2} \times K \times ( {x_{1}}^{2} - {x_{2} }^{2}) \\ \\ \implies W_{s} = \frac{1}{2} \times 100 \times ( - { ( 0.1})^{2} ) \\ \\ \implies W_{s} = 50 \times ( - 0.01) \\ \\ \boxed{ \implies W_{s} = - 0.5J} \\ \\ Now \\ \\ \boxed{ \Delta U = - W_{s} = + 0.5J} \\ \\ therefore \: \: your \: \: answer \: \: is \implies \\ \\ \huge \boxed{ W_{s} = - 0.5 J \: \: \: \: and \: \: \: \: \Delta U = 0.5J}$

$2) \\ \\ \huge\bf{ \mid{ \overline{ \underline{{question} \mid}}}}$

$When \: \: the \: \: spring \: \: is \: \: compressed \: \: \\ by \: \: distance \: \: 'x' \: \: it's \: \: potential \: \: energy \: \: is \: \: U_{1} \\ .It \: \: is \: \: further \: \: compressed \: \: by \: \: distance \: \: \\ '2x' \: \: it's \: \: potential \: \: energy \: \: now \: \: is \: \: U_{2} \\ .Find \: \: U_{1} : U_{2}$

$2) \\ \\ \huge\bf{ \mid{ \overline{ \underline{{answer} \mid}}}} \\ \\ \large{formula :-} \\ \\ W_{spring} = \frac{1}{2} \times K \times ({x_{1} }^{2}-{ x_{2}}^{2} \\ \\ \implies W_{spring} = \frac{1}{2} \times K \times ({x_{1} }^{2}-{ x_{2}}^{2} \\ \\ U_{1} = \frac{1}{2} K ({x}^{2}) ------(1) \\ \\ Now \: \: U_{2} \\ \: \: it \: \: will \: \: x+2x=3x \\ \\ W_{spring} = \frac{1}{2} \times K({(3x)}^{2} - {x}^{2}) \\ \\ W_{spring} = \frac{1}{2} \times K(9{x}^{2} - {x}^{2}) \\ \\ W_{spring} = \frac{1}{2} \times K(8{x}^{2} ) \\ \\ Now, from (1) \\ \\ U_{2} = 8 (\frac{1}{2} K {x}^{2}) \\ \\ \implies U_{2} = 8 \times U_{1} \\ \\ \huge\boxed{ \frac{U_{1}}{U_{2}} = \frac{1}{8} }$