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A student of mass 63.4 kg, starting at rest, slides down a slide 21.2 m long, tilted at an angle of 26.1° with respect to thehorizontal. If the coefficient of kinetic friction between the student and the slide is 0.113, find the force of kinetic friction, theacceleration, and the speed she is traveling when she reaches the bottom of the slide. (Enter the magnitudes.)HINT(a)the force of kinetic friction (in N) N(b)the acceleration (in m/s) m/s(c)the speed she is traveling (in m/s) m/s

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Answer:

fr = 65.46 N ,   a = 8.74 m / s²  and  vf = 19.25 m / s

Explanation:

We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight

      sin 21.2 = Wx / W

      cos21.2 = Wy / W

      Wx = W sin21.2

      Wy = W cos 21.2

We form Newton's equations

X axis

      Wx -fr = m a

Y Axis

     N- Wy = 0

     N = Wy

     fr = μ N

     fr = μ (W cos 21.2)

     fr = 0.113 63.4 9.8 cos 21.2

     fr = 65.46 N

We replace and calculate the acceleration

     W sin 21.2 - μ W cos 21.2 = m a

     a = g (sin21.2 - μ cos 21.2)

     a = 9.8 (without 21.2 - 0.113 cos 21.2)

     a = 8.74 m / s²

This acceleration is along the slope of the slide, so we can calculate the distance

     d = 21.2 m

     vf² = v₀² + 2 a d

    vf² = 0 + 2 a d

    vf = √(2 8.74 21.2)

    vf = √ (370,576)

    vf = 19.25 m / s