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Vertical spring of force constantkhas a sortof catcher on the upper end, the lower endbeing fixed to a floor. The spring is initiallyneither stretched nor compressed.A massis dropped from some distance straight downonto the spring; upon hitting the catcher itcompresses the spring a distancesbefore com-ing to momentary rest.

In terms of these quantities, what was thekinetic energy of the mass the instant it hitthe spring catcher?

Respuesta :

Answer:

K = mg h

Explanation:

We will solve this exercise through energy conservation

Initial. Highest point

       Em₀ = U = m g (h + x)

Final. When the spring hits

       [tex]Em_{f}[/tex] = K + U = K + mg x

       Em₀ = [tex]Em_{f}[/tex]

       m g (h + x) = K + mgx

       mg h (h + x) –mgx = K

       K = mg h

We can also solve it with the points

Initial. Spring contact point

       Em₀ = K + U = K + m g x

Final. Maximum compression

       [tex]Em_{f}[/tex] = [tex]K_{e}[/tex] = ½ k x²

       Emo =[tex]Em_{f}[/tex]

       K + mg x = ½ k x²

       K = ½ k x² - mg x

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