Question

a 2kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. when the spring is 4cm shorter than it's equilibrium length. the speed of the block is root 17 m/s. the greatest speed of the block is​

Answer 1

Answer:

Spring Energy = 1/2KX^2

Kinetic Energy = 1/2mv^2

At the time given, our initial time for our initial energy, it has some kinetic energy and some spring energy.

At the time of it's greatest speed, our final time or final energy, it will be at equilibrium, x=0 and will have all kinetic energy. Since Energy Initial = Energy Final we can set spring energy + kinetic energy at 4 cm = kinetic energy at 0 cm.

Therefore,

1/2 (80) (.04)^2 +1/2 (.5) (.5)^2 = 1/2 (.5) (v)^2 solve for v and you have your answer also the x needs to be in m not cm

.064 + .0625 = (1/4) v^2

v^2= .506

v= .71 m/s

Explanation:

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Answer 2

Answer:

9 Meter Per Second

Explanation:

Explanation in Pic 2.