A toboggan approaches a snowy hill moving at 12.4 m/s. The coefficients of static and kinetic friction between the snow and the toboggan are 0.400 and 0.290, respectively, and the hill slopes upward at 44.0 degrees above the horizontal. What is the acceleration of the toboggan going up the hill? and the acceleration after it has reached its highest point and it sliding downhill?

a) Acceleration while the toboggan is going uphill = 8.85 m/s²

b) Acceleration of the toboggan is going downhill = 4.76 m/s²

Explanation:

Resolving the normal reaction into vertical and horizontal component

Nₓ = mg sin θ

Nᵧ = mg cos θ

The frictional force on the toboggan = Fr = μmg cos θ (note that μ = μ(k) = the coefficient of kinetic friction)

Doing a Force balance on the x component taken in the axis parallel to the inclined plane.

The net force that has to accelerate the toboggan has to match the frictional force acting downwards of the plane (In the opposite direction to motion) and x-component of the Normal reaction

ma = Nₓ + Fr

ma = (mg sin θ) + (μmg cos θ)

a = g[(sin 44°) + (0.29 cos 44°)

a = 9.8(0.6947 + 0.2086) = 8.85 m/s²

b) While coming downhill,

The frictional force is acting uphill now (In the direction opposite to motion),

The force balance is

ma = (mg sin θ) - (μmg cos θ)

a = g[(sin 44°) - (0.29 cos 44°)]

a = 9.8(0.6947 - 0.2086)

a = 4.76 m/s²

true tan speed = ang vel x radius