Originally Posted by
9665
3) A force of 40N accelerates a 5.0kg block at 6.0m/s squared along a horizontal suface.
a) How large is the frictional force?
b) What is the coefficiant of friction?
We have to look at all the forces in play
First we have frictional force
Then there is the force accelerating the block. Because the block is accelerating, the force used is greater than frictional force.
Remember from above that frictional force is equal to mu * normal force or
Therefore the force applied must be greater than frictional force
F(friction)-F(applied) = -ma , It is minus because the forces run in opposite directions, it really depends on YOUR coordinate system
F(friction) + -40 = 5*0*-6.0
F(friction) = 10N
Now for part B
F(friction) = mu* F(normal)
10 = mu*m*g
10 = mu*5*9.8 (If you had used an acceleration of 6.0 here, you would end up with 0.333.. however, normal force deals only with acceleration in the Y direction)
mu = 0.204
I hope this helps