Answer:
v = 13.19 m / s
Explanation:
This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical
X axis
N = m a
Centripetal acceleration is
a = v² / r
Y Axis
fr -W = 0
fr = W
The force of friction is
fr = μ N
Let's calculate
μ (m v² / r) = mg
μ v² / r = g
v² = g r / μ
v = √ (g r /μ)
v = √ (9.8 11 / 0.62)
v = 13.19 m / s
Mechanical energy is the sum of kinetic energy and potential energy, or E=Ek+Ep. So Ek=(1/2)*m*v² where m is the mass of the object and v is it's velocity. Mp=m*g*h where m is the mass, g=9.81 m/s² and h is the height of the object. So after we input the numbers the total mechanical energy is
E=(1/2)*2.5*(4.5²) + 2.5*9.81*18 = 25.3125 J + 441.45 J = 466.7625 J. The correct answer is E= 466 J.
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
Upward force provided by the branch is 260 N
<u>Explanation:</u>
Given -
Mass of Gibbon, m = 9.3 kg
Length of the branch, r = 0.6 m
Speed of the movement, v = 3.3 m/s
Upward force, T = ?
The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.
Therefore,
F net = T - mg
F net = ma = mv²/r
Thus,
T = mv²/r + mg
T = m ( v²/r + g)
T = 9.3 [ ( 3.3)² / 0.6 + 9.8]
T = 259.9 N ≈ 260 N
Therefore, upward force provided by the branch is 260 N