A horizontal force of 150 N is used to push a 41.5 kg packing crate a distance of 6.10 m on a rough horizontal surface. If the crate moves at constant speed. What is the coefficient of kinetic friction between the crate and the surface? Please show your work. Thank You!!!
2006-11-20 09:32:52 · 4 answers · asked by Tennis2127 2 in Science & Mathematics ➔ Physics
Since the crate is moving at constant speed, acceleration = 0, net force is 0.
Applied Force 150 N --> [41.5 kg mass] <-- Friction F = 150 N
For sliding friction, F(friction) = u N , where u (mu) is the coefficient, and N is the Normal force (imagine an arrow pointing down from the mass).
N = mass x g
2006-11-20 09:51:57 · answer #1 · answered by Pony 2 · 0⤊ 0⤋
At constant speed, the acceleration is zero, and your only force is (a) the one applied, and (b) the one due to friction acting in the opposite direction. Summation of forces, we find the two forces are equal. Since friction is defined as mu*N, where N=9.81*41.5, equate to 150 and solve for mu.
2006-11-20 17:42:04 · answer #2 · answered by Daniel W 3 · 0⤊ 0⤋
Force equals the Normal force times the coefficient of friction. Re-arrange and solve for the coefficient. You know the force. You can find the normal force by the mass of the package. The distance moved is irrelevant.
2006-11-20 17:41:28 · answer #3 · answered by Jack 7 · 0⤊ 0⤋
well for the record...i highly doubt your class is harder than mine..seeing as i'm a physics major...the difference between my question and your question is that i actually worked mine out and you seem to be waiting for someone to work through it for you...i didn't mean any harm just for the record.
2006-11-20 22:35:23 · answer #4 · answered by justsinginrain87 3 · 0⤊ 0⤋