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Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?
A) Object 1 rises to the greater height because it weighs less.
B) Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
C) Object 2 rises to the greater height because it contains more mass.
D) Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
The two objects rise to the same height.

Respuesta :

valeriagonzalez0213

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)

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