CP? A 0.150-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 (?Fig. P8.78?). Find the maximum distance the frame moves downward from its initial equilibrium position.
Solution 82P Step 1: 2 Initial potential energy of the spring during stretching, PE = ½ kx Where, k - force constant of the spring x - change in length of the spring We know that, the PE at maximum stretching = Work done by the object to elongate the spring Work done = Force × Displacement Or, W = mgh Where, m - mass of the object g - Acceleration due to gravity h - change in height Provided, m = 0.150kg and h = 0.04 m So, W = 0.150 × 9.8 m/s × 0.04 m W = 0.0588 J This is the potential energy of the spring during maximum stretching Step 2: When a lump of putty is dropped to the frame, the initial potential energy of the lump of putty when it was at rest, U = mgh Here, m = 0.200 kg, g = 9.8 m/s and h = 30 cm = 0.30 m U = 0.200 × 9.8 m/s × 0.30 m U = 0.588 J So, the energy of the lump of putty will be, 0.588 J