A 20.4 kg box initially at rest is pushed 4.08 m along a rough, horizontal floor with a constant applied horizontal force of 97.0653 N. Acceleration of gravity is 9.8 m/s^2.
If the coeficient of friction between the box an the floor is 0.434, find the work done by the friction in units of J.
Also, Find the final speed of the box in units of m/s.
Help!
2006-10-09 08:59:41 · 3 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
OK. The normal force of the box on the floor is
f_n = 20.4*9.8 = 199.92N If the coefficient of kinetic (moving) friction is .434, then the box 'pushes back' with a force of .434*199.92 = 86.76528N Since it is being pushed along by a constant force of 97.0653N the net force on the box is 97.0653 - 86.7652 = 10.3N (rounded)
Since the box travelled a distance of 4.08 meters, the total work done was 4.08*97.0653 = 396.026424 J. Since the box has mass 20.4 kg and the net force is 10.3N, it has an acceleration of 10.3/20.4 = .505 m/s². Since it travelled 4.08m, it's final velocity was
√(2*.505*4.08) = 2.03 m/s (rounded). It's a damned pity they didn't ask you how many seconds it took (about 4.02) to move those 4.08 m ☺
Now it's **your** turn to get off yer dead αsses and figure out how to really *use* all those equations that are in your book ☺
Doug
2006-10-09 09:24:55 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
This is not important. Go to the pub. Now.
2006-10-09 16:06:49 · answer #2 · answered by Nitrous McBread 2 · 0⤊ 0⤋
Wait I am trying to solve it.
2006-10-09 16:20:33 · answer #3 · answered by SKG R 6 · 0⤊ 0⤋