Achild slides down a snow-covered slope on a sled. at the top of the hill, her mother gives her a push to start her off with a speed of 1.00 m/s . the frictional force acting on the sled is one-fifth of the combined weight of the child and the sled. if she travels for a distance of 25.0 m and her speed at the bottom is 4.00 m/s , calculate the angle that the hill makes with the horizontal.

θ = 13.3º

Explanation:

To solve the problem let's use Newton's law, let's start by defining the coordinate system, the x-axis is parallel to the hill (plane) and the y-axis is perpendicular. In this system we must break down the weight

    sin θ = Wx / W

    cos θ = Wy / W

    Wx = W sinT

    Wy = W sinT

We write Newton's equation

X axis

       Wx-fr = m a

Y Axis

      N-Wy = 0

      N = Wy = mg cosT

Calculous la acceleration with cinematic

      vf² = v₀² + 2 a x

      a = (vf²-v₀²) / 2x

      a = (4² - 1) / 2 25

      a = 0.30 m / s²

They indicate that the friction force is 1/5 of the weight of the sled + child

      fr = 1/5 W

We calculate

    W sin θ -1/5 W = m a

    mg (sin θ -1/5) = ma

    sin θ = a / g + 1/5

    sin θ = 0.30 / 9.8 + 1/5

    θ = sin-1 (0.2306)

     θ = 13.3º