Physics, asked by Anonymous, 2 months ago

A spring of mass 2kg and stiffness 1 KN/m is dropped from a height of 2 m on the ground . calculate the maximum distance through which the spring is compressed​

Answers

Answered by nirman95
16

Given:

A spring of mass 2kg and stiffness 1 KN/m is dropped from a height of 2 m on the ground.

To find:

Max compression?

Calculation:

Let max compression be "x".

Now, we can say:

  • Initial PE = mg(2 + x)

  • Final PE = ½kx²

Now, as per law of conservation of energy:

 \rm \: mg(2 + x) =  \dfrac{1}{2} k {x}^{2}

 \rm \implies \: 2 \times 10 \times (2 + x) =  \dfrac{1}{2}  \times 1000 \times  {x}^{2}

 \rm \implies \: 20 (2 + x) =  500{x}^{2}

 \rm \implies \: 40 + 20x =  500{x}^{2}

 \rm \implies \:  500{x}^{2}  - 20x - 40 = 0

 \rm \implies \:  50{x}^{2}  - 2x - 4= 0

 \rm \implies \:  25{x}^{2}  - x - 2= 0

After solving the quadratic equations:

 \rm \implies \: x = 0.3 \: m

So , max compression of spring is 0.3 metres.

Answered by NewGeneEinstein
7

Answer:

To find:-

Maximum compression

Solution:-

Let it be y

Now,

Initial PE=mg(2+y)

Final PE=1/2ky^2

According to Law of conservation of energy:-

\\ \sf\hookrightarrow mg(2+y)=\dfrac{1}{2}ky^2

\\ \sf\hookrightarrow 2\times 10(2+y)=\dfrac{1}{2}\times 1000\times y^2

\\ \sf\hookrightarrow 20(2+y)=\dfrac{1000}{2}\times y^2

\\ \sf\hookrightarrow 40+20y=500y^2

\\ \sf\hookrightarrow 500y^2-20y-40=0

  • Factoring

\\ \sf\hookrightarrow 25y^2-y-2=0

\\ \sf\hookrightarrow y=\underline{+}0.03m

  • Take y Positive

\\ \sf\hookrightarrow y=0.03m\blue{\bigstar}

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