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

Answer:

$a=5.82\ m/s^2$    

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

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

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

$a=9.8\times \ sin(42)-0.1\times 9.8 cos(42)$

$a=5.82\ m/s^2$

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

Answer 2

Answer:

8.68 m/s^2

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