Question

A 0.167-kg frame, when suspended from a coil spring, stretches the spring 0.0400 m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm.

Find the maximum distance the frame moves downward from its initial equilibrium position.

Answer 1

Hi,

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Initially, The force on the spring = mg = (0.167 kg) x (9.8 m/s²) = 1.6366 N

Length of the spring elongated, X = 0.04 m

Therefore, Spring constant of the spring is given by:

Now we nee to apply Law of conservation of energy.

The initial gravitational potential energy of the clay lump will be converted into potential energy of the spring.

Let x be the length of elongation of the spring,

Therefore, maximum distance the frame moves downward from its initial equilibrium position (X) is given by:

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