Answer:
v = 2.029 m/s
Explanation:
Given
L = 84.0 cm ⇒ R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m
m₁ = 0.600 kg
m₂ = 0.200 kg
g = 9.8 m/s²
u₁ = u₂ = 0 m/s
v₁ = ?
v₂ = ?
Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).
In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.
then
Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J
Ef = Kf + Uf
⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²
⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464
⇒ Ef = Kf + Uf = 0.4*v² - 1.6464
Since
0 = 0.4*v² - 1.6464 ⇒ v = 2.029 m/s
v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.
v₁ = v₂ = v ⇒ ω₁*R₁ = ω₂*R₂ ⇒ ω₁*R = ω₂*R ⇒ ω₁ = ω₂ = ω
⇒ v = ω*R
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
Explanation:
∆x=300 m×2
∆t=1.5 s
v=∆x/∆t → v=2×300/1.5 = 400 m/s
