Respuesta :
Explanation:
(a) Net force acting on the block is as follows.
[tex]F_{net} = -mg Sin (\theta)[/tex]
or, ma = -mg Sin (\theta)[/tex]
a = [tex]-g Sin (\theta)[/tex]
= [tex]-9.8 \times Sin (20^{o})[/tex]
= -3.35 [tex]m/s^{2}[/tex]
According to the kinematic equation of motion,
[tex]v^{2} - v^{2}_{o} = 2as[/tex]
Distance traveled by the block before stopping is as follows.
s = [tex]\frac{v^{2} - v^{2}_{o}}{2a}[/tex]
= [tex]\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}[/tex]
= 21.5 m
According to the kinematic equation of motion,
v = [tex]v_{o} + at[/tex]
0 = [tex]12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t[/tex]
[tex]t_{1}[/tex] = 7.16 sec
Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.
(b) When the block is moving down the inline then net force acting on the block is as follows.
[tex]F_{net} = -mg Sin (\theta)[/tex]
ma = [tex]mg Sin (\theta)[/tex]
a = [tex]g Sin (\theta)[/tex]
= [tex]9.8 m/s^{2} \times Sin (20^{o})[/tex]
= 3.35 [tex]m/s^{2}[/tex]
Kinematics equation of the motion is as follows.
s = [tex]v_{o}t + \frac{1}{2}at^{2}[/tex]
21.5 m = [tex]0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}[/tex]
[tex]t_{2}[/tex] = [tex]\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}[/tex]
= 3.58 sec
Hence, total time taken by the block to return to its starting position is as follows.
t = [tex]t_{1} + t_{2}[/tex]
= 7.16 sec + 3.58 sec
= 10.7 sec
Thus, we can conclude that 10.7 sec time it take to return to its starting position.
