Respuesta :
Answer:
The answer to your question is v₁ = 5.74 m/s
Explanation:
Data
v₁ = ?
h₁ = 0 m
v₂ = 0.75 m/s
h₂ = 1.65 m
g = 9.81 m/s²
Formula
mgh₁ + 1/2mv₁² = mgh₂ + 1/2mv₂²
mass is not consider (if we factor mass, it is cancelled)
gh₁ + 1/2v₁² = gh₂ + 1/2v₂²
Substitution
(9.81)(0) + 1/2v₁² = (9.81)(1.65) + 1/2(0.75)²
Simplification
0 + 1/2v₁² = 16.19 + 0.28
Solve for v₁
1/2v₁² = 16.47
v₁² = 2(16.47)
v₁² = 32.94
Result
v₁ = 5.74 m/s
Minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.
Given here,
v₁ - initial velocity = ?
h₁ - initial height = 0 m
v₂ - final velocity = 0.75 m/s
h₂ - final height = 1.65 m
g - gravitational acceleration = 9.81 m/s²
The speed can be calculated by using the formula,
[tex]\bold { mgh_1 + \dfrac 12 mv_1^2 = mgh_2 + \dfrac 12mv_2^2}[/tex]
factor the mass,
[tex]\bold { gh_1 + \dfrac 12 v_1^2 = gh_2 + \dfrac 12v_2^2}[/tex]
put the values in the formula, and solve it for V1
[tex]\bold { (9.81)(0) + \dfrac 12v_1^2 = (9.81)(1.65) + \dfrac 12(0.75)^2}\\\\\bold { \dfrac 12v_1^2 = 16.19 + 0.28}\\\\\bold { \dfrac 12v_1^2 = 16.47}\\\\ \bold {v_1^2 = 2(16.47)}\\\\\bold {v_1^2= 32.94}\\\\\bold { v_1 = 5.74\ m/s}[/tex]
Therefore, minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.
To know more about kinetic energy,
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