A 20.0kg sled is being pulled across a horizontal surface at a constant velocity

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A 20.0kg sled is being pulled across a horizontal surface at a constant velocity....

Hi. This is a problem on my physics homework that I can't seem to get the (correct) answer to. I would love to be told what I'm doing wrong. (:


This is a problem and what I did:

A 20.0kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0 N and is directed at an angle of 30° above the horizontal. Determine the coefficient of kinetic friction.

m(sled) = 20kg
F(gravity) = 20kg * 9.8 m/s2 = 196 N

F(pull) = 80 N
F(pull,X) = 80sin30° = 40 N
F(pull,Y) = 80cos30° = 69.28 N

F(normal) = 196 N - 69.28 N = 126.72 N

Kinetic Friction = μ * F(normal)

μ = 40 N / 126.72 N = .3157

I looked in the back in the book, and apparently the correct answer is .444. What am I doing wrong?

Thanks!
 
Last edited:

Answers and Replies

  • #2
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Nevermind, I figured out what I did incorrectly.
Lol, I mixed the vertical and horizontal components of the F(pull) by accident.
....

m(sled) = 20kg
F(gravity) = 20kg * 9.8 m/s2 = 196 N

F(pull) = 80 N
F(pull,Y) = 80sin30° = 40 N
F(pull,X) = 80cos30° = 69.28 N

F(normal) = 196 N - 40 N = 156 N

Kinetic Friction = μ * F(normal)

μ = 69.28 N / 156 N = .4441
 
Last edited:

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