a 10kg weight is being pushed along a level surface at 3 m/s, with a force of 14.7n. what's the coeficient of friction? @Eyad He's doin it again :P

Question

a 10kg weight is being pushed along a level surface at 3 m/s, with a force of 14.7n.  what's the coeficient of friction? @Eyad He's doin it again :P

anonymous · Answer

F = Normal * coefficient.  So 14.7 = (10*9.8) * x
then you solve for x to be about 0.15.  So the friction coefficient is 0.15.  This is assuming that the block is on earth

blacksteel · Answer

If the speed of the weight is not changing, that means that the total force acting on the block is 0n.
Therefore, the magnitude of the frictional force acting on the block is 14.7n, to counteract the force pushing it. Frictional force is given by normal force N times coefficient of friction u, N*u. Therefore, N*u = 14.7. N is given by mass times acceleration due to gravity, m*g - assuming we're on Earth, g = 9.8 m/s^2, so u = 14.7/(9.8*10) = 14.7/98 = 0.15

lifeisadangerousgame · Answer

Thanks guys!