3. A tree house has a vertical “fire pole” of smooth metal, designed for quick exits.
A child of mass 35.7kg slides down the pole with constant acceleration, starting
from rest. The pole is 3.10 m high. The journey to the ground takes 2.00 s.
(a) What is the magnitude of the downward acceleration of the child?
(b) What is the magnitude of the upward force of friction exerted by the pole
on the child?
(a) To find the acceleration of the child, we first need to find the final velocity. We can use the following kinematic equation:
vf = vi + at
where:
vf = final velocity (0 m/s as the child comes to rest at the bottom)
vi = initial velocity (0 m/s as the child starts from rest)
a = acceleration
t = time taken (2.00 s)
0 = 0 + a(2.00)
a = 0 m/s^2
The magnitude of the downward acceleration of the child is 0 m/s^2.
(b) To find the upward force of friction exerted by the pole on the child, we can use Newton's second law:
ΣF = ma
where:
ΣF = sum of forces
m = mass of the child (35.7 kg)
a = acceleration (0 m/s^2)
The only vertical forces acting on the child are its weight (mg) and the frictional force (f):
ΣF = mg - f = ma
Rearranging the equation:
f = mg - ma
= m(g - a)
= (35.7 kg)(9.8 m/s^2 - 0 m/s^2)
= 350.46 N
The magnitude of the upward force of friction exerted by the pole on the child is 350.46 N.