So the upward force is 400cos30°
The normal force is = mgcos30° + 400sin30°
So than my equation becomes =
400 cos 30° - (343 cos30° + 400 sin30°).µ - 343 sin 30° = 35.0,56
right?
than µ = 0,312
0,5 . 35 1² + 0 = 0,5 . 35 1,8² + 35.9,81.2
The change in KE and PE is = 726J
726J = F.x
F= 726/2 = 363N
And then I'm stuck because I don't see any second force affection the normal force :S
Fpush - (Ffriction + Fgravity) = m . a
400/cos30° - 343 cos 30° µ - 343 sin 30° = 35 . 0,56
µ = 0,91
I find only 1 Force affecting the normal force that is mg cos30°
Hi!
A block with mass 35 kg is pushed upward with a horizontal force of 400N along a slope at an angle of 30 °. The block makes a displacement of 2 meters, where it's speed changes from 1m/s to 1,8m/s. What is the Coefficients Of Friction?
1) 0.312
2) 1.099
3) 0.039
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