A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass sytem is 1.5 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.7 kg block?

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Asked Jun 18, 2019
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A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass sytem is 1.5 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.7 kg block? 

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Expert Answer

Step 1

The mass of the first block =m1=3.2kg.

The elastic potential energy corresponding to block with mass m1=U1=1.5J.

Let m2 be the second mass, and corresponding potential energy be U2.

Write the expression for the potential energy of the spring

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here, k is the spring constant 1 U =-kx 2 (I) x is the compressed distance

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Step 2

 

Write the expression for the force acting on spring.

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F -kx neglect the negative sign] mg -x mg (II) k

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Step 3

Substitute equation (II) in (I) and write the expression for th...

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II) k 2

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