Question

A 56-kg person on skis is going down a hill sloped at 30° from the horizontal. The coefficient of friction between the skis and the snow is 0.16. What would be the magnitude of the acceleration?

Answer 1

3.54 m/s²

Step-by-step explanation:

mass, m = 56 kg

inclination, θ = 30°

coefficient of friction, μ = 0.16

As the person moves down

So, the acceleration is

a = g Sinθ - μ g Cos θ

a = 9.8 ( Sin 30 - 0.16 Cos 30)

a = 9.8 (0.5 - 0.139)

a = 3.54 m/s²

Thus, the acceleration is 3.54 m/s².

Answer 2

Net acceleration,
`a=3.55\ m/s^2`

Step-by-step explanation:

It is given that,

mass of the person, m = 56 kg

Inclination with the horizontal,
`\theta=30^(\circ)`

The coefficient of friction between the skis and the snow is 0.16

Let
`a_g` is the acceleration due to gravity on the object,

`a_g=g\ sin\theta`

`a_g=9.8\ sin(30)`

`a_g=4.9\ m/s^2`

Let
`a_f` is the acceleration of the normal reaction. It is given by:

`a_f=\mu g\ cos\theta`

`a_f=0.16* 9.8* cos(30)`

`a_f=1.35\ m/s^2`

Let a is the net acceleration of the person. It is equal to,

`a=a_g-a_f`

`a=4.9-1.35`

`a=3.55\ m/s^2`

So, the magnitude of net acceleration is
`3.55\ m/s^2`. Hence, this is the required solution.