Answer:
(a) [tex]a_{y}= 1.931m/s^{2}[/tex]
(b) [tex]t=10.478s[/tex]
Explanation:
For Part(a)
Apply ∑Fy=ma
[tex]T-Mg=Ma_{y} \\[/tex]
Solve for a
[tex]a_{y}=\frac{T-Mg}{M}=\frac{2.80mg-Mg}{M} \\a_{y}=\frac{(2.80*575kg-1345kg)}{1345kg}*(9.80m/s^{2} )\\a_{y}= 1.931m/s^{2}[/tex]
For Part(b)
Assume the maximum acceleration (maximum tension upwards is greater than the weight of the chain+boulder)
As the given data
[tex]a_{y}=1.931m/s^{2}\\ y-y_{o}=106m\\v_{oy}=0[/tex]
Using equation of simple motion
[tex]y-y_{o}=v_{oy}t+(1/2)a_{y}t^{2} \\t=\sqrt{\frac{2(y-y_{o})}{a_{y}} }\\ t=\sqrt{\frac{2(106m)}{1.931m/s^{2} } }\\t=10.478s[/tex]
