A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with the horizontal. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between block and ceiling.

0.35

Explanation:

According to Newton's second law;

\sum Fx = ma

Fm - Ff =ma

Fm is the moving force = Wsin theta

Fm = 4(9.8)sin55

Fm = 32.1N

Ff is the frictional force = nmgcos theta

Ff = n(4)(9.8)cos55

Ff = 22.48n

Acceleration a = 6.0m/s²

Substitute the given values into the formula and get the coefficient of friction

32.11-23.48n = 4(6)

32.11-24= 23.48n

8.11 = 23.48

n = 8.11/23.48

n = 0.35

Hence the coefficient of friction is 0.35

b

explanation:

bc its just b