A block is given an initial speed of 3m/s up a 25° incline coefficient of friction= 0.12
A) how far up the plane will it go?
B) how much time elapses before it returns to it's starting point?.

a)   y = 0.459 m , b)   t = 13 s

Explanation:

a) For this exercise we use Newton's second law to find the acceleration of the block

we fix a reference system with the x axis parallel to the plane and the y axis perpendicular to it

X axis

          -Wₓ - fr = m a               (1)

Axis y

          N - = 0

          N = W_{y}

the friction force has the formula

          fr = μ N

let's use trigonometry to find the components of the weight

          sin 25 = Wₓ / W

         Wₓ = W sin 25

          cos 25 = W_{y} / W

          W_{y} = W cos 25

we substitute in 1

          - W sin 25 - μ W cos 25 = m a

          - g (sin 25 - μ cos 25) = a

let's calculate

          a = - 9.8 (sin 25 - 0.12 cos 25)

         a - 0.25 m / s

this is the acceleration on the plane, so we can use the kinematic relations in one dimension

the highest point where the block rises the speed is zero (va = 0)

          v² = v₀² - 2 a y

           y = v₀² / 2g

          y = 3 2 / (2 9.8)

           y = 0.459 m

this is the distance the cantes block travels to stop

  b) the time of the entire journey is

               y = v₀ t - ½ a t²

the point where the body recesses is y = 0

            0 = (vo - ½ a t) t

whose solutions are

           t = 0

           0 = vo - ½ a t

           t = 2vo / a

           t = 2 3 / 0.459

           t = 13 s

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