1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N
4) The net force on the sled acting in the vertical direction is made up of
• the sled's weight, 585 N, pointing downward
• the vertical component of the applied force, 20 N, pointing upward
• the normal force, with magnitude n, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ F = n + 20 N - 585 N = 0  ==>  n = 565 N
5) The net force in the horizontal direction consists of
• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)
• kinetic friction, with magnitude f, pointing in the opposite and negative direction
By Newton's second law,
∑ F = 34.6 N - f = 0  ==>  f ≈ 34.6 N
Now if µ is the coefficient of kinetic friction, then
f = µn  ==>  µ = f/n = (34.6 N) / (565 N) ≈ 0.0613
The component of the force is the effective part of that force in that direction.
The component of the force is the effective part of that force in that direction.
1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N
2) The vertical component of the force = 40 N sin 30 degrees = 20 N
3) The magnitude of the gravitational force = mg cos 30 degrees  = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N
4) The normal force = 551.7 N
5) The coefficient of friction = F/R = Â 40 N /551.7 N = 0.07
Learn more about component of a force:https://brainly.com/question/15529350
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