Question

A small block of mass m slides along the inner, frictionless surface of a vertical circular track.

At point A the magnitude of the block’s acceleration is 2g.

What is the magnitude of the normal force on the block at point B?
a) 4mg
b) 5mg
c) 6mg
d) 7mg
e) 8mg

What is the magnitude of the normal force on the block at point C?
a) 4mg
b) 5mg
c) 6mg
d) 7mg
e) 8mg

Answer 1

a) 4mg

c) 6mg

Step-by-step 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).