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Place the single weight with a known mass on the spring and release it. Eventually, the weight will come to rest at an equilibrium position, with the spring somewhat stretched compared to its original (unweighted) length. At this point, the upward force of the spring balances the force of gravity on the weight. With the weight in its equilibrium position, how does the amount the spring is stretched depend on the value of the weight's mass?

Respuesta :

lcmendozaf

Answer:

X=m*g/K

Explanation:

Since the elastic force of the spring is balancing the force of gravity:

Fe = m*g

Now, on the spring by Hook's law, the magnitude of the elastic force will be:

Fe = K*X     where K is the elastic constant of the spring and X is the distance the spring is strectches measured from its original lenght to its current length.

Replacing this value:

K*X = m*g   Solving for X:

X = m*g/K    This value is directly proportional to the object's weight and inversely proportional to the spring's constant.

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