Question

A 65.0 kg skier slides down a37.2° slope with uk = 0.107.What is the skier's acceleration?(Unit = m/s2)Enter

Answer 1

`a=5.08m/s^2`

Step-by-step explanation

Parameters given

Mass of skier = 65 kg

The angle of slope = 37.2°

Coefficient of friction = 0.107

First, let us make a sketch of the problem:

We need to find the skier's acceleration.

To do this, we have to apply the resultant of forces acting on the skier as he slides down.

We have that the resultant of forces is:

`\begin{gathered} Fg_x-Fr=ma_x \\ F_N-Fg_y=0 \end{gathered}`

where FN = normal force acting on the body

Fg = weight

Fr = frictional force acting on the body.

The x and y components of the weight of the body are given as:

`\begin{gathered} Fg_y=mg\cos (37.2) \\ Fg_x=mg\sin (37.2) \end{gathered}`

where m = mass

The frictional force acting on the body is:

`Fr=\mu\cdot mg\cos (37.2)`

Using the horizontal component, make a the subject of the formula:

`a=(Fg_x-Fr)/(m)`

Now, solving for a using the parameters given:

`\begin{gathered} a=(mg\sin (37.2)-\mu mg\cos (37.2))/(m) \\ a=g\sin (37.2)-\mu gcos(37.2) \\ a=(9.8\sin 37.2)-(0.107\cdot9.8\cdot\cos (37.2) \\ a=5.92-0.84 \\ a=5.08m/s^2 \end{gathered}`

That is the value of the acceleration.