Question

A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providing a steady force of 148 N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 75.7 kg, and the coefficient of kinetic friction between the snow and the runners of the sled is 0.299. How much time is required for the sled to travel down a 124-m slope, starting from rest?

Answer 1

Answer:

 travel time is 7.58 s

Explanation:

We must know the acceleration of the sled and the girl, for this we must use Newton's second law

      F = m a

Since this is a vector equation, let's separate it into its xy components, the best selection of the coordinate system is the X axis parallel to the plane and the Y axis perpendicular to it.

In the attached you can see a free body diagram of this situation, let's write Newton's equations on each axis

X axis

      Wx + F - fr = m a            (1)

Axis y

     N- Wy = 0

     N = Wy

Let's use trigonometry to find the weight components

    Wx = W sin θ

    Wy = W cos θ

In addition, the formula for friction force is

    fr = n N

Let's clear the acceleration of equation 1

   

    a = (W sin θ + F - m N) / m

    a = (mg sin θ + F - m mg cos θ) / m

    a = g (sin θ -m cos θ) + F / m

Let's replace and calculate

    a = 9.8 (sin 30 - 0.299 cos 30) + 148 / 75.7

    a = 2.36 + 1.96

    a = 4,317 m / s²

We already have the acceleration, the speed is zero to leave the rest (Vo = 0) and give the length that runs 124 m, let's use the kinematic equations

   x = v₀ t + ½ a t²

   x = ½ a t²

   t² = 2x / a

  t = √ 2x / a

 

Let's calculate

  t = √( 2 124 / 4,317) = √ 57.45

  t = 7.58 s