Question

Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall be deflected 45 mm downwards from the equilibrium position and released. At what point in time does the weight have the highest speed?

Answer 1

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds