A 10 kg block sits on a flat surface whose μs=0.60 and whose μk is 0.40

a. What horizontal force is required to get the block move?

b.If we continue to apply the same force as in part a. what will the blocks acceleration be?

In the answer:
force=.6*mg to move
net force=ma
first force-.6mg=ma solve for a

My question:

what is the first force??... should i subtract it to .6mg?

thanks.

(a) F=F(fr) = μ(s) •N=

= μ(s) •m•g=0.6•10•9.8=58.8 N

(b) ma=F-F1(fr) =
=F- μ(k) •N= F- μ(k) •m•g,
a = (F/m) - μ(k)•g= (58.8/10) – 0.4•9.8 =
=5.88-3.92= 1.96 m/s²

In this context, the "first force" referred to is the force required to get the block to move, which is also known as the static friction force. To determine this force, you need to use the equation:

force = μs * mass * gravity

Where:
μs is the coefficient of static friction (given as 0.60)
mass is the mass of the block (given as 10 kg)
gravity is the acceleration due to gravity (approximately 9.8 m/s²)

So, the first force is:
force = 0.60 * 10 kg * 9.8 m/s²

After calculating this value, you can subtract it from the equation for the net force (force - 0.6*mg) to find the acceleration of the block.