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Viefleur [7K]
3 years ago
8

A 0.250 kg block on a vertical spring with a spring constant of 5.00 x 103 N/m is pushed downward compressing the spring 0.100 m

. When released, the block leaves the spring and travels upward vertically. How high does it rise above its point of release
Physics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant k = 5 \times 10^{3} \frac{N}{m}

Compression x = 0.10 m

Mass of block m = 0.250 kg

Here spring potential energy converted into potential energy,

   mgh = \frac{1}{2} kx^{2}

For finding at what height it rise,

  0.250 \times 9.8 \times h = \frac{1}{2}  \times 5 \times 10^{3} \times (0.10) ^{2}                ( ∵ g = 9.8 \frac{m}{s^{2} } )

  h = 10.20 m

Therefore, the height at point of release is 10.20 m

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