Respuesta :
it is the same, force diagram is the same, acceleration depends on forces. the only thing different is the initial velocity.
Answer:
[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
Explanation:
Always draw a diagram!
Up the incline:
[tex]Fr_{max}[/tex] = maximum friction
[tex]Fr_{max}[/tex] = μk
k = R = mg.cos(45) = mg.sin(45)
Resolution of forces parallel to the slope:
F (Fp in the diagram) = force of propulsion
g = gravity
[tex]F - Fr_{max} = ma_{i}[/tex]
[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]
Down the decline:
Resolution of forces:
[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]
[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]
Then, find the ratio:
[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]
Potentially, there is no need to consider F in this situation, in which case:
[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
