Part A A toboggan approaches a snowy hill moving at 12.1 m/s2 . The coefficients

Question

Part A
A toboggan approaches a snowy hill moving at 12.1 m/s2 . The coefficients of static and kinetic friction between the snow and the toboggan are 0.420 and 0.270, respectively, and the hill slopes upward at 42.0 degrees above the horizontal.

+Find the acceleration of the toboggan as it is going up the hill.
Assume +x axis directed up the hill.

+Find the acceleration of the toboggan after it has reached its highest point and is sliding down the hill.
Assume +x axis directed up the hill.


Part B
A toboggan approaches a snowy hill moving at 12.6 m/s2. The coefficients of static and kinetic friction between the snow and the toboggan are 0.410 and 0.260, respectively, and the hill slopes upward at 44.0 degrees above the horizontal.

+Find the acceleration of the toboggan as it is going up the hill.
Assume +x axis directed up the hill.

+Find the acceleration of the toboggan after it has reached its highest point and is sliding down the hill.
Assume +x axis directed up the hill.

Explanation / Answer

As the tobaggan is moving up the hill, it experiences no forces up the hill, and two forces back down the hill: gravity's component down the hill, and kinetic friction. The component of gravity down the hill is given by the sine of the angle:

(F_{grav} = mgsin heta)

The force of friction will be

(F_{friction} = mu_{k}N = mu_{k}mgcos heta)

Keep in mind, the normal force is balanced by the cosine component of gravity in the case of an inclined plane. Now, putting these together, the acceleration is

(a = F_{net}/m = (-mgsin heta - mu_{k}mgcos heta)/m = -gsin heta - mu_{k}gcos heta)

Plugging in g = 9.81 m/s^2 and mu_k = 0.26, we get

(a = -8.56 m/s^{2})

Now, when it is sliding down the hill, the force of gravity and the force of kinetic friction have the same formula, but while gravity still points down the hill, kinetic friction is now pointing up the hill. Therefore, the acceleration is

(a = F_{net}/m = (mu_{k}mgcos heta - mgsin heta)/m = mu_{k}gcos heta - gsin heta)

Plugging in our values, we get

(a = -4.83 m/s^{2})

Hope this helps.

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