A child pulls a 3.00-kg sled across level ground at constant velocity with a light rope that makes an angle 34.0° 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)0.156 B)0.188 C)0.0441 D)0.0851 E)0.103

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Quizplus · Accepted Answer

The coefficient of friction is given by:$$\mu_k = \frac{F_f}{N}$$where:* $F_f$ is the force of friction* $N$ is the normal forceThe force of friction is equal to the tension in the rope:$$F_f = T = 5.00 	ext{ N}$$The normal force is equal to the weight of the sled:$$N = mg = (3.00 	ext{ kg})(9.81 	ext{ m/s}^2) = 29.43 	ext{ N}$$Therefore, the coefficient of friction is:$$\mu_k = \frac{F_f}{N} = \frac{5.00 	ext{ N}}{29.43 	ext{ N}} = 0.169$$The closest answer choice is A) 0.156.

A Child Pulls a 3