Answer:
10.32m/s
Explanation:
F = ma
T -mg sinθ-f =ma
We can conclude that T = 300 newtons by taking a freebody of the cable. So we need to determine only the frictional force.
f(n) = 0
mgcosθ = N = 0
N = mgcosθ
T - mg sinθ - u mgcosθ = ma
a = T - mg (sinθ + u cosθ) / m
[tex]a = \frac{300 - 26(9.8)(\sin 30^0 + 0.3 \cos 30^0)}{20}[/tex]
a = 5.32m/s^2
Use constant acceleration kinematics:
[tex]v_f^2=V_i^2+2ad\\V_f^2 = \sqrt{2ad}[/tex]
= [tex]= \sqrt{2\times 5.32\times10} \\\\=10.32m/s[/tex]
