I keep getting different answers, so I need some help with this:
A force of 40N accelerates a 5.0 kg box at 6.0 m/s*2 along a horizontal surface. How large is the frictional force?
2007-03-03 17:14:24 · 3 answers · asked by skybluu 2 in Education & Reference ➔ Homework Help
Since the object is experiencing a net forward force of 30N (5*6m/s^2=30) and the force applied to it is 40N, the friction must result in a force of 10N in the opposite direction. That is 40N-10N=30NIncidentally, the coefficient of friction is .2, since .2*(5*10m/s^2=weight)=10N.
2007-03-03 17:23:22 · answer #1 · answered by bloggerdude2005 5 · 1⤊ 0⤋
lets assume that the frictional force if F, it is always in the opposite direction of the motion. there fore apply Newton's second law of motion in the direction of motion
ie
40 - F = 5*6
F = 10N
2007-03-04 01:49:58 · answer #2 · answered by san 3 · 0⤊ 0⤋
Whoah what is this? Physics?
2007-03-04 01:27:31 · answer #3 · answered by ray91_91 2 · 0⤊ 0⤋