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

1.25377 m/s²

Step-by-step 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=(gsin\theta-a)/(gcos\theta)\\\Rightarrow \mu=(9.81* sin5-0.4)/(9.81* 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* sin10-0.04655* 9.81* 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