A force of 20N is needed to start moving an object of mass 4kg but it needs only 16N to keep the object moving at a constant velocity. What is the magnitude of the force of

a) kinetic friction b) limiting friction?

F_f = \mu R

Where Ff is the frictional force, mu either the coefficient of static friction or that of kinetic friction and R the normal reaction of the ground on the object.

a) F_f = \mu_1 R = \mu_1 mg

Ff = 20 N
m = 4 kg
g = 9.8 m/s^2

b) F_f = \mu_2 R = \mu_2 mg

Ff = 16 N
m = 4 kg
g = 9.8 m/s^2

Solve for mu1 and mu2

The formula that is given to me to use is Uk = fk/Fn and Us = fs/Fn (I don't know how to do that u sign) and I don't know where to put the kg into the formula.

If you look carefully at Unknown008's equations, you'll see they're actually the same as yours, just transposed a little and with different names for the variables.

Your equations use Uk (coefficient of kinetic friction), whereas Unknown's use \mu_2. Yours use Us (coefficient of static friction), whereas Unknown's use \mu_1. You use fk and fs, whereas unknown called both forces F_f, and you used Fn (normal force), whereas unknown used R to represent the same thing.

Note that you can easily transpose your equations to look the same:

Uk = fk/Fn --> fk = Uk * Fn

The transposed equation is the same as F_f = \mu_2 R

As unknown said, you use the object's mass (4 kg) when computing the normal force, Fn. In this case, the normal force is the force of gravity (i.e. the object's weight), which is just m*g (4 kg * 9.8 m/s^2 = 39.2 N).

So regardless of whether you use Unknown008's formulas or the ones given to you, just plug in the numbers and solve for Uk and Us (or \mu_2 and \mu_1).

Does it make more sense now?

Wow thank you :):):):) this really helped :)