A child 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? A) 0.161 B) 0.188 C) 0.0441 D) 0.0851 E) 0.103

Question

Quizplus · Accepted Answer

The horizontal component of the tension is T*cos(30°) = 5.00 N * (√3/2) = 4.33 N. The normal force is reduced by the vertical component of the tension, T*sin(30°) = 5.00 N * (1/2) = 2.50 N. The normal force is then (3.00 kg * 9.81 m/s²) - 2.50 N = 27.93 N. The frictional force equals the horizontal component of the tension, so μ * 27.93 N = 4.33 N, giving μ = 4.33 N / 27.93 N = 0.155, which is closest to 0.161.

A Child Pulls a 3