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

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft