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A man pulls a 30.0 kg crate across level ground at constant velocity with a light...

A man pulls a 30.0 kg crate across level ground at constant velocity with a light rope that makes an angle of 20.0° above horizontal. The tension in the rope is 40.0 N. 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. 0.134

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Answer #1

Find the normal force acting on the box and use that to find the friction force. Since the velocity is constant, the acceleration will be zero. Use this in the equation of motion using Newton’s law of motion to find the coefficient of friction as shown belowNot Tsino = mg => No + (40N)sin20 = (301949-8m/s2) » No = 280-32N eq. of motion is, Tcoso - f = Ma » Troso - No = 0 {as a=o d

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