A 10kg box rests ona ramp that is laying flat. The coefficient of static friction ( µs ) is 0.50, and the coefficient of kinetic friction ( µk) is 0.30.....
what is the force (Fp) necessary to keep the box sliding at a CONSTANT VELOCITY?
I'm not really sure how to work this problem...this is the only thing i could come up with, but I dont think its right...could check this plz?
∑Fy = Fn - Fg = 0 , therefore
Fn = Fg = mg = 10kg*9.8m/s2 = 98N
∑Fx = Fp - Ffk = 0, therefore
Fp = Ffk = µk(Fn) = 0.30 (98N) = 29.4N
The reason why I think there might be something wrong with this is b/c i calculated the max static friction (Ffs) to be 49N...Im really confused could someone plz help me?
2006-10-22 09:08:52 · 1 answers · asked by Sir Excalibur 2 in Science & Mathematics ➔ Physics
Hey, not to worry. You did this exactly correctly. The answer is indeed 29.4 N. You're just noticing that the force of kinetic friction is less than the force of static friction. That's typical, and not a red flag on your answer.
2006-10-27 07:41:10 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋