According the attached picture so we can assume that :
The coefficient of friction wood on wood are:
static friction μs = 0.5
dynamic friction μk = 0.2
a)The answer is:
F= 588.6 N
The explanation:
The maximum force you can exert horizontally on the crate without moving it is equal:
Fsmax = μs * M * g
when μs is the static friction
and M i she mass
g is the gravitational force by substituted:
Fsmax = 0.5 * 120Kg * 9.81 m/s2
= 588.6 N
when F= Fsmax
So the answer is F = 588.6 N
b)The answer is:
a= 2.943 m/s2
The explanation:
Fsmax is the force pushing the block , and there is a force of friction trying to stop the block .
So Fk = μk * N
when Fk = force of kinetic friction
μk = coefficient of kinetic friction of wood on wood
N= Normal force by substitution:
Fk= 0.2 * 120k *9.81 m/s2 = 235.44 N
so F = Fsmax - Fk
588.6 - 235.44 = 353.16 N
when F = m*a
so 353.16 = 120Kg * a
so a ( the acceleration) = 353.16/120kg
= 2.943 m/s2