The coefficient of kinetic friction between the block and the table to the nearest hundredth is 0.07
According to Newton's second law of motion:
[tex]\sum F_x=ma_x[/tex]
[tex]\sum F_x[/tex] is the sum of forces
m is the mass
[tex]a_x[/tex] is the acceleration
[tex]\sum F_x = F_m-Ff[/tex]
Fm is the moving force
Ff is the frictional force
The formula becomes
[tex]F_m-F_f=ma_x\\F_m - \mu R =ma_x\\F_m-\mu mg = ma_x[/tex]
Given the following
m = 3kg
[tex]a_x[/tex] = 3m/s²
Fm = 11N
g = 9.8m/s²
Substitute the given values into the formula:
[tex]11-\mu(3\times 9.8)=3 \times 3\\11-\mu(29.4)=9\\-\mu(29.4)=9-11\\-\mu(29.4)=-2\\29.4\mu=2\\\mu=\frac{2}{29.4}\\\mu= 0.068\\\mu \approx 0.07[/tex]
Hence the coefficient of kinetic friction between the block and the table to the nearest hundredth is 0.07
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