A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4 cm longer than its equilibrium length, the speed of the block is 0.50 m/s. The greatest speed of the block is
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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
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
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0
Explanation:
first find total energy, and at maximum speed the total energy is kinetic energy. then use kinetic energy relation to find out the maximum speed...!!!
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