A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the surface were frictionless. b. How large is the kinetic friction force? cWhat is the coefficient of kinetic friction?

a = 8 m/s^2, Ffriction = 10 N, μk = 0.205

Explanation:

a. Force = Mass*Acceleration,

(since you didn't add the units..."5 block"for the mass, I will assume it to be in kg, per SI units)

40 N = 5 kg*acceleration,

a = 40/5 = 8 m/s^2

b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.

Fapplied - Ffriction = m*a,

40 - Ffriction = 5*6,

40 - Ffriction = 30,

Ffriction = 40 - 30 = 10 N

c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".

10 = μk*Fnormal (Fnormal = m*g = 5*9.8)

10 = μk*49,

μk=10/49 ≈ 0.205

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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