Achild pulls a 3.00-kg sled across level ground at constant velocity with a light rope that makes an angle 30.0° above horizontal. the tension in the rope is 5.00 n. assuming the acceleration of gravity is 9.81 m/s2, what is the coefficient of friction between the sled and the ground?

μ = 0.161

Explanation:

Since the tension on the rope (F) is being applied at an angle (θ), it should be decomposed into a vertical and a horizontal component.

The vertical component of the tension on the rope (F*sin(θ)) is pulling the sled upwards, decreasing the normal force.

Therefore, the normal force can be calculated as:

The horizontal component of the tension on the rope (F*cos(θ)) is the force being applied to move the sled and that should equal the coefficient of friction multiplied by the normal force.

the coefficient of friction between the sled and the ground is 0.161

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)