A toboggan approaches a snowy hill moving at 11.7 m/s. The coefficients of static and kinetic friction between the snow and the toboggan are 0.48 and 0.34, respectively, and the hill slopes upward at 40.0 ∘ 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?

The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².

Explanation:

Given that,

Speed = 11.7 m/s

Coefficients of static friction = 0.48

Coefficients of kinetic friction = 0.34

Angle = 40.0°

(a). When the toboggan moves up hill, then

We need to calculate the acceleration

Using formula of acceleration

Put the value into the formula

(b). When the toboggan moves up hill, then

We need to calculate the acceleration

Using formula of acceleration

Put the value into the formula

Hence, The acceleration of the toboggan going up and down the hill is 8.85 m/s² and 3.74 m/s².

+15-35=-20

going in opposite directions