Inactive Tutor answered • 09/21/20
Hi there!
First off, friction is caused by the tiny imperfections on two surfaces, like going over bumps in the road. Friction operates off of the normal force, which is the force that you drew pushing against the ground at 20 N. The formula for the force of friction in general is:
Ff = µ * Fn
where Ff is the force of friction,
µ (mu, the greek letter for m) is the coefficient of kinetic friction (kinetic vs. static or rolling, because it's sliding),
and Fn is the normal force. Generally, normal force can be found by:
Fn = mg cos Θ
where m is the mass
g is gravity
and Θ is the angle between the slope of the surface and the object's weight pulling straight down (cos is the cosine function).
In this case, the surface is not slanted and so is 90° from the weight pulling straight down, so cos Θ = 1 and can be ignored here because mg * 1 = mg. I'm showing you these extra details now because you will likely learn this in the future.
For your problem, I will assume that as shown, the object is experiencing an external rightward force of 8 N, so the force of friction is thus 4 N pushing to the left. Thus, the magnitude is 4 N. If my assumption is wrong, then 8 N is the magnitude.
Next, we should rearrange our friction equation to solve for µ and then plug the value we pick in for Ff:
Ff = µmg
µ = mg / Ff
= 2.0 kg * 9.8 m/s² / Ff
≈ 20 N / Ff
For Ff = 4 N, µ = 5.0 (unitless)
For Ff = 8 N, µ = 2.5 (unitless)
Either way the force of friction is (please remember to clarify on your homework with your given problem), the net force is 4 N to the right because 4 N to the left pushes back on the 8 N that much. We know that force is mass * acceleration:
F = ma
So, we can solve for acceleration:
a = F / m,
and we know the force and the mass, so we can plug them in:
a = Fnet / m
= (8 N - 4 N) / 2.0 kg
= 4 N / 2.0 kg
= 2.0 m/s²
Hope this helps! Please reach out with any further questions.
