Respuesta :
Answer:
Case A requires more force than Case B
Explanation:
For this exercise, we propose Newton's second law, we assume that the box is pushed on the x-axis
X axis
F -fr = m aₓ
F = fr + m aₓ
Y Axis
N- W = m [tex]a_{y}[/tex]
N = m a_{y} + m g
N = m (a_{y} + g)
The friction force has as an expression
fr = μ N
We substitute
F = μ m (g + a_{y}) + m aₓ
Let's apply this equation to our case s
Case A. The elevator moves upward so ay is positive
F₁ = μ m (g + a_{y}) + m aₓ
Case B. The elevator is move down so that ay is negative
F₂ = μ m (g - a_{y}) + m aₓ
We see that F₁ > F₂
Case A requires more force than Case B
For the force to be the same the friction must be zero, which is an ideal chao since there is always some friction
