The block's speed when two springs used to launch the same block is 0.71 m/s.
The problems not complete, look at the attachment to see the picture. In picture (2) two springs paralleled. The constant for two paralleled springs
k = k₁ + k₂
k = 25 + 25
k = 50 N/m
When the springs compressed, the conservation energy law applies.
K.E block + P.E springs = K.E' block + P.E' springs
- The block initially stop.
Kinetic energy for the block before launching is zero.
K.E block = 0 - The potential energy for spring before launching
P.E springs = 0.5 × k × Δx² - The kinetic energy for the block after launching
K.E' block = 0.5 × m × v² - The potential energy for spring after launching is zero.
P.E' springs = 0 - Δx = 10 cm = 0.1 m
- m = 1 kg
0 + 0.5 × k × Δx² = 0.5 × m × v² + 0
k × Δx² = m × v²
50 × 0.1² = 1 × v²
0.5 = v²
[tex]v = \sqrt{0.5}[/tex]
v = 0.71 m/s
Learn more about conservation energy law here: https://brainly.com/question/11550087
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