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
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
To find the coefficient of friction for the surface, we can use the formula:
Coefficient of friction = Force of friction / Normal force
In the first scenario, the force required to push the block at a constant velocity is given as 2.5 N. This force is equal to the force of friction opposing the motion.
Therefore, the coefficient of friction for the surface in the first scenario is:
Coefficient of friction = 2.5 N / 6.8 N = 0.3676
Now, let's move on to the second scenario where a weight W is placed on the block. In this case, the total force needed to push both the block and the weight at a constant velocity is given as 4.6 N. Again, this force is equal to the force of friction opposing the motion.
To find the weight W, we need to subtract the force required to push only the block (2.5 N) from the total force required to push both (4.6 N):
W = Total force - Force required to push the block
W = 4.6 N - 2.5 N
W = 2.1 N
Therefore, the weight W placed on the block is 2.1 N.