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Yes, you should get 0.05 m. The equation is 1/2mv^2 = 1/2kx^2, so x = v/sqrt(k). Plugging in the values, you get x = 5/sqrt(200) = 0.05 m.
I have a question involving the spring costant:
A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring? The answer is supposed to be in m.
I used 1/2mv^2=1/2KX^2, the answer I get is .5m which doesn't seem right.
Thank you
The word equilibrium is confusing here, so I am assuming as you did that the equilibrium length for a horizontal spring is its stretched length.
You used the correct relationship: max PE equals starting KE.
then if follows
x= v/sqrt k
I don't get your answer, I get less.
did u get .05m? i don't why i am getting that now
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