2.5 N = (weight) x (kinetic friction coefficient)
Solve for the kinetic friction coefficient.
In part 2, the total weight being pushed increased by a factor 4.6/2.5 = 1.84
The total weight being pushed must therefore have increased by the same factor, to 12.5 N.
The added weight must have been 12.5 - 6.8 = __ N
A block weighing 6.8 N requires a force of
2.5 N to push it along at constant velocity.
What is the coefficient of friction for the
surface?
A weight W is now placed on the block and
4.6 N is needed to push them both at constant
velocity.
What is the weight W of
1 answer
1 answer