Respuesta :
Answer:
a) at start Ek(₀) =0 after 1 meter Ek(₁) = 54.1 [J] change in Ek 0 54.1 [J]
b) At 1 meter down Ek(₀) = 54.1 [J] 2 meter Ek(₁) = 108.19 [J] change in Ek
in relation to the start point is 108.19 [J] an in relation to previous one meter is 54.1 [J]
Explanation:
Kinetic energy is Ek = 1/2*m*v²
and 1 J = Kg* m² /s²
a) When dropping the object ( V(₀) = 0 and then the kinetic energy is 0
since Ek₀ = 1/2 * m* 0 = 0
at the distance of 1 meter, according to the equation
Vf² = V₀² + 2*a*d V₀ = 0 then
Vf² = 2*g*1 ⇒ Vf² = 2*9.8 m/s² * 1 m ⇒ Vf² = 19.6 m²/s²
Vf = 4.43 m/s
And Ek(₁) = 1/2* m* Vf² ⇒ Ek(₁) = 1/2*5.52*19.6 [J]
Ek(₁) = 54.1 [J]
And the change in kinetic energy is 54.1 [J]
b) Again Vf² = V₀² + 2*a*d but now we have 2 meters
Then Vf² = 2*9.8 m/s² * 2 m ⇒ Vf² = 39.2 m²/s² ⇒ Vf = 6.26 m/s
And
Ek(₂) = 1/2*5.52* 39.2 [J] ⇒ Ek(₂) = 108.19 [J]
