Question

A block that weighs 200 N is initially at rest on a horizontal surface where the coefficient of static friction is 0.500 and the coefficient of kinetic friction is not yet known.

A) What is the gravitational force on this block?
B) What is the mass of the block?
C) What is the normal force on this block?
D) What is the smallest force needed to start moving the block?
E) If you were pushing the block with the force you found in C, what force would the block exert on you?
F) If you were pushing the block with the force you found in C, and the block began to accelerate at 1.96 meters per second squared, what is the net force on the block?
G) If the block accelerates at 1.96 meters per second squared, what is the coefficient of kinetic friction?

Answer 1

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

Step-by-step 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