G: st = 38.92-17.59 FN = (90kg ) (9.8 m/s²) = 21.35 = 882w R: a =?, MK = ? A: FF ↑ Ev=Fg [40 Kay + 50kg |= 40 ку FA=SOON

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Chapter6: Applications Of Newton's Laws
Section: Chapter Questions
Problem 114AP: A 15-kg sled is pulled across a horizontal, snow-covered surface by a force appiled to a rope at 30...
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I was able to do the given, required and analysis. Can you please show me a solution
G: At = 38.92-17·59 FN = (90kg ) (9.8 m³5²)
= 21.35
= 882 N
R: a =?, MK = ?
A:
F
↑Fo=Fg
показ
+ 50kg
1 = 90kg
J Fg
FA=500N
Transcribed Image Text:G: At = 38.92-17·59 FN = (90kg ) (9.8 m³5²) = 21.35 = 882 N R: a =?, MK = ? A: F ↑Fo=Fg показ + 50kg 1 = 90kg J Fg FA=500N
Speed
0.0 m/s
50 kg
Let me know IF there is enough information to calculate the coefficient of kinetic friction.
I can only tell you that the applied force is 500 N.
IF
you can calculate the coefficient of kinetic friction, show me using GRASS.
IF you do not have enough information, make an assumption that would help you solve for the
coefficient of kinetic friction.
40 kg
Transcribed Image Text:Speed 0.0 m/s 50 kg Let me know IF there is enough information to calculate the coefficient of kinetic friction. I can only tell you that the applied force is 500 N. IF you can calculate the coefficient of kinetic friction, show me using GRASS. IF you do not have enough information, make an assumption that would help you solve for the coefficient of kinetic friction. 40 kg
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