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:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
Answer:

Explanation:
<u>Accelerated Motion</u>
The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

Or, equivalently

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

The negative sing of a indicates there is deceleration or decreasing speed