Question

A 0.175-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 (Figure 1).

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

Express your answer with the appropriate units.

remained 2 rights to answer the question. please do correct!!!!!!!!!!!!!!!!!!!!!

Answer 1

Let's find the spring constant: F = mg = kd

k = mg / d = 0.150kg * 9.8m/s² / 0.040m = 36.75 N/m

After the putty hits it:

Let's say the spring stretches "x," and that our GPE is measured from the point of maximum stretch.

Then our putty starts out at height x + 0.300m

and our frame starts out at height x

and our spring starts out with stretch 0.040 m.

All of this potential energy gets converted into spring PE, so

(0.150kg * x + 0.200kg * (x + 0.300m)) * 9.8m/s² + ½ * 36.75N/m * (0.040m)²

= ½ * 36.75N/m * (x + 0.040m)²

This solves to x = 0.24 m (= 24 cm)