How Does the Coefficient of Static Friction Affect a Backpack on a Table?

[bc64846]

a 52 N backpack is laying on a table with a spring attached to one side of it where force is applied and the spring stretches. The backpack begins to slide when the spring (k= 150 N/m) stretches by 2.50 cm. What is the coefficient of static friction between the backpack and the table?

i set up a freebody diagram, but i guess I'm just confused about the weight given in N. I have only had problems dealing with kg. I set up adding all forces in the x and y direction and got:

-F + fs = ma(x direct)
-Fg + FN = ma(y direct)

FN is normal force, fs is static force

I got the F by using F= -kx and got 3.75N

 

[phlegmy]

if the spring stretches 2.5cm =.025m so the force is
150N/m * .025 = 3.75n

now this is the force reqired to just overcome the static friction

so assuming the table is horizontal the reaction force is 52N

now 3.75N= cof of frict * 52N

soooooooooooooooo......

 

[m2003]

.

The weight given in N is not a problem, as it is simply a measurement of force. In this case, the weight of the backpack is 52 N, meaning that it exerts a force of 52 N on the table due to gravity. This force must be balanced by the normal force from the table, which is equal in magnitude but opposite in direction. Therefore, the normal force is also 52 N.

To find the coefficient of static friction, we can use the equation fs = μsFN, where μs is the coefficient of static friction. We already know the value of fs (3.75 N) and FN (52 N), so we can plug these values into the equation to solve for μs.

3.75 N = μs * 52 N

μs = 3.75 N / 52 N = 0.072

Therefore, the coefficient of static friction between the backpack and the table is 0.072. This means that the force needed to keep the backpack from sliding is only 7.2% of the force exerted by the backpack due to gravity. This low coefficient of friction suggests that the surfaces of the backpack and the table are relatively smooth, allowing for easy sliding.