Question

A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is the acceleration of the same skier when she is moving down a hill with a slope of 10 ∘? Assume the coefficient of kinetic friction is the same in both cases.

Answer 1

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction

$\theta$ = Slope

From Newton's second law

$mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655$

Applying $\mu$ to the above equation and $\theta=10^{\circ}$

$mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2$

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²