Energy in a spring:
E = 0.5 * k * x²
k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m
E = 0.5 * 800 * 0.05² = 1
Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
A) the electrode at which oxidation takes place
