Answer:
fr = 65.46 N
, a = 8.74 m / s² and vf = 19.25 m / s
Explanation:
We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight
sin 21.2 = Wx / W
cos21.2 = Wy / W
Wx = W sin21.2
Wy = W cos 21.2
We form Newton's equations
X axis
Wx -fr = m a
Y Axis
N- Wy = 0
N = Wy
fr = μ N
fr = μ (W cos 21.2)
fr = 0.113 63.4 9.8 cos 21.2
fr = 65.46 N
We replace and calculate the acceleration
W sin 21.2 - μ W cos 21.2 = m a
a = g (sin21.2 - μ cos 21.2)
a = 9.8 (without 21.2 - 0.113 cos 21.2)
a = 8.74 m / s²
This acceleration is along the slope of the slide, so we can calculate the distance
d = 21.2 m
vf² = v₀² + 2 a d
vf² = 0 + 2 a d
vf = √(2 8.74 21.2)
vf = √ (370,576)
vf = 19.25 m / s