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A basket of negligible weight hangs from a vertical spring with force constant k. (a) A mass m is gently lowered into the basket and the spring is allowed to reach equilibrium. I.e. the mass is not released until the spring is in equilibrium. How far is the spring stretched? (b) This time the mass is dropped into the basket. I.e. the mass is released at the level of the basket causing the basket to oscillate. Find the maximum distance that the spring is stretched. (c) If the mass is dropped into the basket, from a height h above the basket, by how much will the spring stretch at its maximum elongation?

Respuesta :

Answer:

a)   x = m g / k , b) x= 0, c) x =√( mg/k)  √2h

Explanation:

a) The spring must comply with Hooke's law

         F = k x

When we put a dough in the basket, we must fulfill the equilibrium equation

         F - W = 0

           

        k x = m g

        x = m g / k

This is the distance the spring lengthens

b) In this case the mass is released and the system oscillates, the maximum amplitude of the spring is given by the energy carried by the mass when released

If the mass is loose from a height h, its energy is

       Em₀ = U = m g h

       Emf = Ke = 1/2 k x²

       1/2 k x² = mg h

       x =√( mg/k)  √2h

Therefore the amplitude must be equal to h

         x = 0

With the body released at a height h very close to hard the amplitude should be very close to zero

c) x = h

Amplitude equals height for conservation of mechanical energy

       x =√( mg/k)  √2h

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