Answer:
The speed of the spider is v = (2g*L*(1-cosθ))^1/2
Explanation:
using the energy conservation equation we have to:
Ek1 + Ep1 = Ek2 + Ep2
where
Ek1 = kinetic energy = 0
Ep1 = potential energy = m*g*L*cosθ
Ek2 = (m*v^2)/2
Ep2 = m*g*L
Replacing, we have:
0 - m*g*L*cosθ = (m*v^2)/2 - m*g*L
(m*v^2)/2 = m*g*L*(1-cosθ)
v^2 = 2g*L*(1-cosθ)
v = (2g*L*(1-cosθ))^1/2
Take the derivative to find the velocity of the object:
The object stops when 
:
so the answer is E.
Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
