Question

A spring/mass system is shown in four different states:

1. The spring hangs vertically in equilibrium with no mass at its end.
2. A 2.6 kg mass hangs from the spring in equilibrium, stretching the spring b=5.7 cm.
3. The mass is pulled down a distance c and released.
4. The mass is at the reference height moving with a speed of v=8.4 m/s.

How far down (c) was the mass pulled before it was released? (Hint: you can calculate the spring constant with the given information)
(include units with answer)

Answer 1

At reference point of this oscillation, Kinetic energy (K.E.) must be equal to Potential Energy (P.E.)

Now, K.E. = m x v² / 2

and, P.E. = kd² / 2 ,

where, m = mass = 2.6 kg , v = speed at reference point = 8.4 m/s, d = distance by which the mass was pulled,

k = spring constant = gravitational force by the mass m / displacement (b), b = 5.7 cm = 0.057 m

= (2.6 kg x 9.8 m/s²) / 0.057 m = 447 N/m

Hence, m x v² / 2 = kd² / 2 gives,

2.6 x 8.4² = 447 x d²

or, d = 0.64 m = 64 cm

Hence, the mass was pulled by 64 cm