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Norma-Jean [14]
2 years ago
12

The 3kg object in Figure is released from

Physics
1 answer:
Arlecino [84]2 years ago
4 0

(a) The compression of the spring is 0.074 m.

(b) The potential energy of block will be converted to elastic potential energy of the spring when the block comes to rest.

<h3>Compression of the spring [ x ]</h3>

The compression of the spring is calculated as follows;

F = kx

mg = kx

x = mg/k

where;

  • m is mass of the object
  • k is the spring constant

x = (3 x 9.8) / (400)

x = 0.074 m

<h3>Motion of the block</h3>

Based on the principle of conservation of energy, the potential energy of block will be converted to elastic potential energy of the spring when the block comes to rest.

Learn more about elastic potential energy here: brainly.com/question/25996974

#SPJ1

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\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

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  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

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  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

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