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atroni [7]
3 years ago
14

The force required to stretch a Hooke’s-law

Physics
1 answer:
Setler [38]3 years ago
5 0

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


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The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

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<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

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