Question

2. A 2.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.60, and the coefficient of kinetic friction is µk = 0.40. A horizontal force, F, is applied to the block. As F is increased, the block begins moving. Describe how the force of friction varies as F increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of F―before the block starts moving, at the point it starts moving, and after it is moving.
Could someone please explain this question and the steps?

Answer 1

I really want to understand how to do this! Thank you :)

Answer 2

Well, since it is a force question, you're best helper will be Newton's second law.
\(\Sigma F=m \Sigma a\)
Now, you'll have to analyse the situations.
First the block does not move even if a force is exerted.
As the force increases until one point, the static friction force suddenly gives way and turns into kinetic friction because the block is moving.
So first case is \(\Sigma F=F-F_fs=0\)
so,\(F=F_fs\)
Second case is ,\(\Sigma F=F-F_fk=ma\)
Do you follow?

Answer 3

The formula for static friction is \(F_fs \le \mu_s F_N\) while the formula for static friction is \(F_fk=\mu_k F_N\).

Answer 4

Yes! That makes more sense now. However, what does the E symbol represent?
By "ma" do you mean mass * acceleration?

Answer 5

The E symbol is the greek letter "Sigma", which represents the total sum.
Yup! :) I believe you've got it :)