Question

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

Answer 1

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