Answer:
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
