Question
Asked Jun 18, 2019
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A vertical spring with a spring constant of 480 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.0 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?

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Expert Answer

Step 1

The spring constant of the spring = 480 N/m.

The mass of the block =0.30 kg.

The compressed distance = 2 cm.

Consider the figure given. Initially block is at height h, when it is dropped towards the spring, the spring compressed to a distance of 2 cm.

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m 0.30kg x - 2 cm k 480N/m Lo0g0g0040000000

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Step 2

Hence at height h, block has a potential energy mgh as it dropped down it is completely converted to kinetic energy, which is responsible for a compression of x distance.

According to low of conservation of energy

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here, m is the mass, k is the spring constant g is the acceleration due gravity kx mgh (I) x is the compressed distance 2 h is the distance between uncompressed srping and block

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Step 3

Rearrange equation (I) to obtain an expression fo...

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1 kx mgh 2 [substitute, 480 N/m for k 0.30kg for m 9.8m/s2 for g | 2x102 m for x in equation (II 1 kx h= 2 mg (II) 1 (480N/m)(2x10m) 2 (0.30 kg) (9.8m/s2) =326.53x 10 x102 cm =3.27 cm

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