Answer:
it will go up along the inclined plane by d = 9.62 m
Explanation:
As we know that puck is moving upward along the slide
then the net force opposite to its speed is given as
[tex]F_{net} = - mgsin\theta - \mu mgcos\theta[/tex]
so here deceleration is given as
[tex]a = -g(sin\theta + \mu cos\theta)[/tex]
now plug in all values in it
[tex]a = -9.81(sin20 + 0.2cos20)[/tex]
[tex]a = -5.2 m/s^2[/tex]
now the distance covered by the puck along the plane is given as
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 10^2 = 2(-5.2)d[/tex]
[tex]d = 9.62 m[/tex]
