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?

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

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

higher potential

explanation:

because it is in a high speed so it must move into a of high potential