A 60.0 kg skier starts from rest at the top of a hill with a 30.0 slope. She reaches the bottom of the slope 4.00 s later. If there is a constant 72.0 N friction force that resists her motion, how long is the hill?

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To solve, the acceleration of the skier along the incline should be determined. To do so, draw the FBD of the skier. (Refer to the attached figure.)

The forces acting on the skier are the friction force, normal force, and force due to gravity. Since the motion is along the...

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To solve, the acceleration of the skier along the incline should be determined. To do so, draw the FBD of the skier. (Refer to the attached figure.)

The forces acting on the skier are the friction force, normal force, and force due to gravity. Since the motion is along the inclined only (x-axis), consider only the forces that have x-components. Base on the FBD, these are the friction force and force due to gravity only. The friction force opposes the motion of the skier, while the x-component of gravitational force is going in the same direction of that of the skier. So, the equation needed to solve for the acceleration of the skier is:

`F_g_x - F_k = ma`

`mgsin(theta) - F_k = ma`.

Plugging in the known values, this becomes:

`60*9.8*sin(30^o) - 72 = 60*a`.

Solving for a, its value will be:

`a=(60*9.8*sin(30^o)-72)/60`

`a=3.7`.

So, the acceleration of the skier as it goes down the slope is `3.7m/s^2` .

To solve for the distance traveled by the skier, apply the kinematics equation for motion with constant acceleration.

`Delta x = v_i t + 1/2at^2`

Since the skier starts from rest, its initial velocity is zero. So, plugging in the known values, this formula becomes:

`Delta x = 0*4 + 1/2*3.7*4^2`

`Delta x =29.6 `.

Therefore, the length of the side of the hill is 29.6m.

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