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A board is hung from two springs, as shown in the figure above, with the board in equilibrium. Each spring has a spring constant of 10,000 N/m. How far will each spring stretch when a person of mass 50 kg sits on the board and the board again comes to equilibrium?

Respuesta :

The distance at which both the springs stretch prior comes to equilibrium is 1/40m.

Calculation of the distance:

Since we know that

x = mg/2k

= 50*9.8/2*10,000

= 0.0245m

= 1/40m

Hence, we can conclude that The distance at which both the springs stretch prior comes to equilibrium is 1/40m.

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