Answer:
E)The two objects rise to the same height : h₁=h₂ =( v²) / (2g)
Explanation:
With coefficient of kinetic friction, μk =0:
We apply the principle of energy conservation :
E₀ = Ef Formula (1)
K₀+U₀ = Kf + Uf Formula (2)
K = (1/2) *m*v² Formula (3)
U = m*g*h Formula (4)
Where:
E₀:
Initial total energy (J)
Ef: Final total energy (J)
K₀: Initial kinetic energy (J)
U₀: Initial potential energy (J)
Kf: Final kinetic energy (J)
Uf:
Final kinetic energy (J)
v : speed (m/s)
m: mass (kg)
h : hight (m)
Known data
v₁=v₂=v
m₁=m
m₂ =2m
μk =0 : coefficient of kinetic friction
Problem development
We apply formulas (1), (2), (3) and (4) for the first object ,(1):
E₀₁ = Ef₁
K₀₁+U₀₁ = Kf₁ + Uf₁ U₀₁=0 , Kf₁ =0 , m₁=m
(1/2) *m*v²+0 =0+m*g*h₁ :We eliminate m,then,
(1/2) *v²=g*h₁
h₁ =( v²) /(2g)
We apply formulas (1), (2), (3) and (4) for the second object ,(2):
E₀₂ = Ef₂
K₀₂+U₀₂ = Kf₂ + Uf₂ U₀₂=0 , Kf₂ =0 , m₂=2m
(1/2) *2m*v²+0 =0+2m*g*h₂ :We eliminate m,then,
(1/2) *2*v² =2*g*h₂ : We divide by 2 on both sides of the equation,then,
(1/2) *v²=g*h₂
h₂ =( v²) / (2g)
