Answer:
a) 4mg
c) 6mg
Explanation:
Given
aca = 2g
We can see the pic shown in order to understand the question
a) We can apply
Ea = Eb
⇒ Ka + Ua = Kb + Ub
⇒ 0.5*m*va² + m*g*ha = 0.5*m*vb² + m*g*hb
⇒ 0.5*m*va² + m*g*(2R) = 0.5*m*vb² + m*g*R
If ac = v²/R ⇒ v² = ac*R
then
0.5*m*(aca*R) + m*g*(2R) = 0.5*m*vb² + m*g*R
⇒ 0.5*(2g*R) + 2gR = 0.5*vb² + g*R
⇒ vb² = 4gR
The Normal force on the block at point B is
N = Fc = m*acb
⇒ N = m*(vb²/R) = m*(4gR/R) = 4mg
The answer is the option a).
b) We can use the equation
Eb = Ec
⇒ Kb + Ub = Kc + Uc
⇒ 0.5*m*vb² + m*g*hb = 0.5*m*vc² + m*g*hc
⇒ 0.5*m*(4gR) + m*g*R = 0.5*m*vc² + m*g*0
⇒ 2gR + gR = 0.5*vc²
⇒ vc² = 6gR
we use the equation
N = m*(vc²/R) = m*(6gR/R) = 6mg
The answer is the option c).
