The conservation of energy allows finding the result of the velocity of the body 2 when it reaches the ground is:
v = 6.32 m / s
Given parameters
To find
The conservation of energy is one of the most important principles of physics, it establishes that if there is no friction force, the mechanical energy is conserved at all points, where the mechanical energy is the sum of the kinetic energy plus the different potential energies.
Let's write the energy at two points before we drop the system.
Em₀ = U₁ + U₂
Where potential energy is
U = m g y
They indicate that body 1 is on the ground and body 2 is at a height .
y₂ = 8 m
Em₀ = m₂ g y₂
Let's write the energy of the system for when body 2 is reaching the ground
[tex]Em_f[/tex] = K₂ + K₁ + U₂ + U₁
At this moment body 2 is reaching the ground, its height is zero and the height of body 1 is y₁ = 8 m. Since the two bodies are joined by the rope, they must have the same speed..
[tex]Em_f[/tex] = ½ (m₁ + m₂) v² + m₁ g y₁
As there is no friction, the energy is conserved.
Em₀ = Em_f
m₂ g y₂ = ½ (m₁ + m₂) v² + m₁ g y₁
The rope is inextensible, the heights are equal.
y₁ = y₂ = y = 8.0 m
v² = [tex]\frac{m2-m1}{m2+m1} \ 2g y[/tex]
Let's calculate.
[tex]v^2 =\frac{22.9-13.6}{ 22.9+13.6} \ 2 \ 9.8 \ 8 \\ v = \sqrt{39.95}[/tex]
v = 6.32 m / s
In conclusion using the conservation of energy we can find the result of the velocity when the body 2 reaches the ground is:
v = 6.32 m / s
Learn more here: brainly.com/question/14688403
