Question

Calculate the magnitude of the acceleration of a sled that is heading down a 42° slope (one that makes an angle of 42° with the horizontal). The coefficient of kinetic friction on ice is μk = 0.100. (Take the x direction to be parallel to and down the slope and the y direction to be perpendicular to the slope.)

Answer 1

8.68 m/s^2

Step-by-step explanation:

angle of inclination, θ = 42°

coefficient of friction, μ = 0.1

g = 9.8 m/s^2

the acceleration of the sled is

a = g Sin θ - μ g Cos θ

a = 9.8 (Sin 42 - 0.1 x Cos 42)

a = 9.8 (0.96 - 0.074)

a = 8.68 m/s^2

Answer 2

`a=5.82\ m/s^2`

Step-by-step explanation:

It is given that,

Angle of inclination of the sled,
`\theta=42^(\circ)`

The coefficient of kinetic friction on ice is,
`\mu_k=0.1`

Using the free body diagram, the magnitude of the acceleration of a sled that is heading down is given by :

`ma=mg\ sin\theta-f`

f is the frictional force

`ma=mg\ sin\theta-\mu mg cos\theta`

`a=g\ sin\theta-\mu g cos\theta`

`a=9.8* \ sin(42)-0.1* 9.8 cos(42)`

`a=5.82\ m/s^2`

So, the magnitude of the acceleration of a sled is
`5.82\ m/s^2`. hence, this is the required solution.