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A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest.

After approaching half the distance \left ( \frac{x}{2} \right ) from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms -1. The total initial energy of the spring is :

  • Option 1)

    1.5 J

  • Option 2)

    0.6 J

  • Option 3)

    0.3 J

  • Option 4)

    0.8 J

 

Answers (1)

best_answer

As we discussed in

Potential Energy stored in the spring -

U= \frac{1}{2}\: kx^{2}

- wherein

K=spring \: constant

x= elongation or compression of spring from natural position

 

 Initial energy of spring

    E_{0}=\frac{1}{2}kx^{2}                              .......(1)

After approaching half the distance its energy will be 

    E_{1}=\frac{1}{2}k.(\frac{x}{2})^{2}=\frac{1}{8}kx^{2}
This momentarily comes to result. Its kinetic energy at that moment =\frac{1}{2}kx^{2}-\frac{1}{8}kx^{2}=\frac{3}{8}kx^{2}

After coming to rest, total kinetic energy is transfered to 2^{nd} mass.

K.E. of second mass =\frac{1}{2}\times(0.1)\times3^{2} =\frac{3}{8}kx^{2}

0.45 J = =\frac{3}{4}(\frac{1}{2}\times kx^{2})

\Rightarrow E_{0}=0.60 J


Option 1)

1.5 J

This is an incorrect option.

Option 2)

0.6 J

This is the correct option.

Option 3)

0.3 J

This is an incorrect option.

Option 4)

0.8 J

This is an incorrect option.

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