Answer:
A) W= 200N, B) m = 20.4 kg
, C) N = 200 N
, D) F = 100 N
,
E) F_reaction = 200 N
, F) F _net = 40 N
Explanation:
This exercise can be solved using Newton's second law
A) The gravitational force is the weight of the block
Fg = W = m g
W = 200 N
B) m = W / g
m = 200 / 9.8
m = 20.4 kg
C) Normal is the reaction of the floor to the weight of the block
N-W = 0
N = W
N = 200 N
D) we write Newton's second law on the x-axis
F - fr = 0
F = fr
the friction force equation is
fr = μ_s N
fr = μ_s W
subtitute
F = 0.50 200
F = 100 N
E) As the forces in the natural are in pairs, by Newton's third law or law of action and reaction, the block responds with a force of equal magnitude, but opposite direction
F_reaction = 200 N
F and G) We write Newton's second law
F - fr = m a
fr = N = μ_k mg
F - μ_k m g = m a
μ_k = (F - ma) / mg
μ_k = (200 - 20.4 1.96) / 200
μ_k = 0.8
In general, the coefficient of kinetic friction is lower than the static one
the net force is
F_net = F -fr = F - μ_k W
F_net = 200 - 0.8 200
F _net = 40 N
