1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) <em>g</em> = (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 <em>n</em>, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ <em>F</em> = <em>n</em> + 20 N - 585 N = 0 ==> <em>n</em> = 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 <em>f</em>, pointing in the opposite and negative direction
By Newton's second law,
∑ <em>F</em> = 34.6 N - <em>f</em> = 0 ==> <em>f</em> ≈ 34.6 N
Now if <em>µ</em> is the coefficient of kinetic friction, then
<em>f</em> =<em> µn</em> ==> <em>µ</em> = <em>f</em>/<em>n</em> = (34.6 N) / (565 N) ≈ 0.0613