Question

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?

Answer 1

Answer:

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

$a=g(\sin\theta+\mu_{k}\cos\theta)$

Put the value into the formula

$a=9.8(\sin40+0.34\times\cos40)$

$a=8.85\ m/s^2$

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

We need to calculate the acceleration

Using formula of acceleration

$a=g(\sin\theta-\mu_{k}\cos\theta)$

Put the value into the formula

$a=9.8(\sin40-0.34\times\cos40)$

$a=3.74\ m/s^2$

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